Class 9: Simultaneous Linear Equations – Exercise 7.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Solve the pair of simultaneous equations using the substitution method.

(i) $\displaystyle 2x+3y = 12 \text{ and } 3x+2y=18$

(ii) $\displaystyle \frac{b}{a} x + \frac{a}{b} y = a^2 + b^2 \text{ and } x+y = 2ab$

(iii) $\displaystyle \frac{x}{6} = y-6 \text{ and } \frac{3x}{4} = 1+y$

(iv) $\displaystyle \frac{x}{2} + \frac{2y}{3} = -1 \text{ and } x - \frac{y}{3} = 3$

(v) $\displaystyle 9-(x-4)= y + 7 \text{ and } 2(x+y) = 4-3y$

Answer:

(i) $\displaystyle 2x+3y = 12 \text{ and } 3x+2y=18$

The given system of equation is:

$\displaystyle 2x+3y = 12$ … … … … … (i)

$\displaystyle 3x+2y=18$ … … … … … (ii)

From equation (i), we get

$\displaystyle x = \frac{1}{2} (12-3y) \Rightarrow x = 6 - \frac{3}{2} y$

$\displaystyle \text{Substituting } x = 6 - \frac{3}{2} y$ in equation (ii), we get

$\displaystyle 3 ( 6 - \frac{3}{2} y) +2y=18$

$\displaystyle \Rightarrow 18 - \frac{9}{2} y = 18$

$\displaystyle \frac{9}{2} y = 0 \Rightarrow y = 0$

$\displaystyle \text{Putting } y = 0 \text{ in } x = 6 - \frac{3}{2} y \text{ we get } x = 6$

$\displaystyle \text{Hence the solution of the given system of equations is } x =6, y = 0$

$\displaystyle \\$

(ii) $\displaystyle \frac{b}{a} x + \frac{a}{b} y = a^2 + b^2 \text{ and } x+y = 2ab$

The given system of equation is:

$\displaystyle \frac{b}{a} x + \frac{a}{b} y = a^2 + b^2$ … … … … … (i)

$\displaystyle x+y = 2ab$ … … … … … (ii)

From equation (ii), we get

$\displaystyle x = (2ab-y)$

$\displaystyle \text{Substituting } x = (2ab-y)$ in equation (i), we get

$\displaystyle \frac{b}{a} (2ab-y) + \frac{a}{b} y = a^2 + b^2$

$\displaystyle \Rightarrow 2b^2 - \frac{b}{a} y + + \frac{a}{b} y = a^2 + b^2$

$\displaystyle \Rightarrow \frac{a^2-b^2}{ab} y = a^2 + b^2$

$\displaystyle \Rightarrow y = ab$

$\displaystyle \text{Putting } y = ab \text{ in } x = (2ab-y) y \text{ we get } x = ab$

$\displaystyle \text{Hence the solution of the given system of equations is } x =ab, y = ab$

$\displaystyle \\$

(iii) $\displaystyle \frac{x}{6} = y-6 \text{ and } \frac{3x}{4} = 1+y$

The given system of equation is:

$\displaystyle \frac{x}{6} = y-6$ … … … … … (i)

$\displaystyle \frac{3x}{4} = 1+y$ … … … … … (ii)

From equation (i), we get

$\displaystyle y = \frac{x}{6} + 6$

$\displaystyle \text{Substituting } y = \frac{x}{6} + 6$ in equation (ii), we get

$\displaystyle \frac{3}{4} x = 1+ \frac{x}{6} + 6$

$\displaystyle \Rightarrow \Big( \frac{3}{4} - \frac{1}{6} \Big) x = 7$

$\displaystyle \Rightarrow x = 12$

$\displaystyle \text{Putting } x = 12 \text{ in } y = \frac{12}{6} + 6 \text{ we get } y = 8$

$\displaystyle \text{Hence the solution of the given system of equations is } x =12, y = 8$

$\displaystyle \\$

(iv) $\displaystyle \frac{x}{2} + \frac{2y}{3} = -1 \text{ and } x - \frac{y}{3} = 3$

The given system of equation is:

$\displaystyle \frac{x}{2} + \frac{2y}{3} = -1$ … … … … … (i)

$\displaystyle x - \frac{y}{3} = 3$ … … … … … (ii)

From equation (ii), we get

$\displaystyle x = \frac{y}{3} +3$

$\displaystyle \text{Substituting } x = \frac{y}{3} +3$ in equation (i), we get

$\displaystyle \frac{1}{2}\Big( \frac{y}{3} + 3 \Big) + \frac{2y}{3} = -1$

$\displaystyle \Rightarrow \frac{y}{6} + \frac{3}{2} + \frac{2y}{3} = -1$

$\displaystyle \Rightarrow y \Big( \frac{1}{6} + \frac{2}{3} \Big) = -1 - \frac{3}{2}$

$\displaystyle \Rightarrow y \frac{5}{6} = - \frac{5}{2}$

$\displaystyle \Rightarrow y = -3$

$\displaystyle \text{Putting } y = -3 \text{ we get } x = 3 + \frac{-3}{3} = 2$

$\displaystyle \text{Hence the solution of the given system of equations is } x =2, y = -3$

$\displaystyle \\$

(v) $\displaystyle 9-(x-4)= y + 7 \text{ and } 2(x+y) = 4-3y$

The given system of equation is:

$\displaystyle 9-(x-4)= y + 7 \Rightarrow x+y = 6$ … … … … … (i)

$\displaystyle 2(x+y) = 4-3y \Rightarrow 2x+5y =4$ … … … … … (ii)

From equation (i), we get

$\displaystyle x = 6-y$

$\displaystyle \text{Substituting } x = 6-y$ in equation (ii), we get

$\displaystyle 2(6-y)+5y = 4$

$\displaystyle \Rightarrow 12 - 2y + 5y = 4$

$\displaystyle \Rightarrow y = - \frac{8}{3}$

$\displaystyle \text{Putting } y = - \frac{8}{3} \text{ in } x = 6 - ( \frac{- 8}{3} ) \text{ we get } x = \frac{26}{3}$

$\displaystyle \text{Hence the solution of the given system of equations is } x = \frac{26}{3}$ , $\displaystyle y = - \frac{8}{3}$

$\displaystyle \\$

Question 2: Solve the system of equations using the method of elimination.

(i) $\displaystyle \frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \text{ and } \frac{3}{x} + \frac{2}{y} =0$

(ii) $\displaystyle 8v-3u=5uv \text{ and } 6v-5u=-2uv$

(iii) $\displaystyle \frac{1}{2(2x+3y)} + \frac{12}{7(3x-2y)} = \frac{1}{2} \text{ and } \frac{7}{2x+3y} + \frac{4}{3x-2y} = 2$

(iv) $\displaystyle 0.4x+0.3y=1.7 \text{ and } 0.7x-0.2y=0.8$

(v) $\displaystyle \frac{x}{2} + y = 0.8 \text{ and } \frac{7}{x+\frac{y}{2}} = 10$

(vi) $\displaystyle \frac{x}{3} + \frac{y}{4} = 11 \text{ and } \frac{5x}{6} - \frac{y}{3} = -7$

(vii) $\displaystyle 3x- \frac{y+7}{11} + 2 = 10 \text{ and } 2y + \frac{x+11}{7} = 10$

(viii) $\displaystyle \frac{1}{2x} - \frac{1}{y} = -1 \text{ and } \frac{1}{x} + \frac{1}{2y} = 8$

(ix) $\displaystyle \frac{6}{x+y} = \frac{7}{x-y} + 3 \text{ and } \frac{1}{2(x+y)} = \frac{1}{3(x-y)}$

(x) $\displaystyle \frac{5}{x+y} - \frac{2}{x-y} = -1 \text{ and } \frac{15}{x+y} + \frac{7}{x-y} = 10$

(xi) $\displaystyle \frac{1}{2(x+2y)} + \frac{5}{3(3x-2y)} = -\frac{3}{2} \text{ and } \frac{5}{4(x+2y)} - \frac{3}{5(3x-2y)} = \frac{61}{60}$
(xii) $\displaystyle \frac{2}{3x+2y} + \frac{3}{3x-2y} = \frac{17}{5} \text{ and } \frac{5}{3x+2y} + \frac{1}{3x-2y} = 2$

(xiii) $\displaystyle \frac{5}{x+1} - \frac{2}{y-1} = \frac{1}{2} \text{ and } \frac{10}{x+1} + \frac{2}{y-1} = \frac{5}{2}$

(xiv) $\displaystyle x-y+z=4$ , $\displaystyle x-2y-2z=9 \text{ and } 2x+y+3z=1$

(xv) $\displaystyle x-y+z=4$ , $\displaystyle x+y+z=2 \text{ and } 2x+y-3z=0$

(xvi) $\displaystyle \frac{10}{x+y} + \frac{2}{x-y} = 4 \text{ and } \frac{15}{x+y} - \frac{9}{x-y} = -2$

(xvii) $\displaystyle \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \text{ and } \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = - \frac{1}{8}$

Answer:

(i) $\displaystyle \frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \text{ and } \frac{3}{x} + \frac{2}{y} =0$

$\displaystyle \text{Taking } \frac{1}{x} = u \text{ and } \frac{1}{y} = v \text{ , the given system of equations becomes }$

$\displaystyle 12u + 4v = 2$ … … … … … (i)

$\displaystyle 3u+2v= 0$ … … … … … (ii)

Now multiplying equation (ii) by $\displaystyle 4$ and subtracting it from equation (i) we get

$\displaystyle 12u+4v=1$

$\displaystyle \underline{ (-) \hspace{0.5cm} 12u+8v=0}$

$\displaystyle -4v=1$

$\displaystyle \Rightarrow v = - \frac{1}{4}$

$\displaystyle \text{Substituting } v = - \frac{1}{4}$ in (ii) we get

$\displaystyle 3u+ 2 (- \frac{1}{4} ) = 0 \Rightarrow u = \frac{1}{6}$

$\displaystyle \text{Therefore } x = \frac{1}{u} = 6 \text{ and } y = \frac{1}{v} = -4$

$\displaystyle \\$

(ii) $\displaystyle 8v-3u=5uv \text{ and } 6v-5u=-2uv$

If we put $\displaystyle v=0$ in either of the equations, we get $\displaystyle u = 0$ . Similarly if we put $\displaystyle u=0$ in either of the equations, we get $\displaystyle v = 0$ . $\displaystyle \text{Therefore } u=0 \text{ and } v=0$ is one of the solutions that satisfies the two equation. However, for $\displaystyle u \neq 0 \text{ and } v \neq 0$ , we follow the following approach.

Dividing each of the given equations by uv we get:

$\displaystyle \frac{8}{u} - \frac{3}{v} = 5$ … … … … … (i)

$\displaystyle \frac{6}{u} - \frac{5}{v} = -2$ … … … … … (ii)

$\displaystyle \text{Taking } \frac{1}{u} = x \text{ and } \frac{1}{v} = y \text{ , the given system of equations becomes }$

$\displaystyle 8x - 3y = 5$ … … … … … (iii)

$\displaystyle 6x - 5y= -2$ … … … … … (iv)

Now multiplying equation( (i) by $\displaystyle 3$ and equation (ii) by $\displaystyle 4$ we get

$\displaystyle 24x - 9y = 15$ … … … … … (v)

$\displaystyle 24x - 20y= -8$ … … … … … (vi)

Subtracting (vi) from (v) we get

$\displaystyle 24x - 9y = 15$

$\displaystyle \underline{ (-) \hspace{0.5cm} 24x - 20y= -8}$

$\displaystyle 11y = 23$

$\displaystyle \Rightarrow y = \frac{23}{11}$

$\displaystyle \text{Substituting } y = \frac{23}{11}$ in (iii) we get

$\displaystyle 8x - 3( \frac{23}{11} ) = 5 \Rightarrow x = \frac{31}{22}$

$\displaystyle \text{Therefore } u = \frac{1}{x} = \frac{22}{31} \text{ and } v = \frac{1}{y} = \frac{11}{23}$

$\displaystyle \\$

(iii) $\displaystyle \frac{1}{2(2x+3y)} + \frac{12}{7(3x-2y)} = \frac{1}{2} \text{ and } \frac{7}{2x+3y} + \frac{4}{3x-2y} = 2$

$\displaystyle \text{Taking } \frac{1}{2x+3y} = u \text{ and } \frac{1}{3x-2y} = v \text{ , the given system of equations becomes }$

$\displaystyle \frac{1}{2} u + \frac{12}{7} v = \frac{1}{2} \Rightarrow 7u + 24v = 7$ … … … … … (i)

$\displaystyle 7u + 4v= 2$ … … … … … (ii)

Subtracting (ii) from (1) we get

$\displaystyle 7u + 24v = 7$

$\displaystyle \underline{ (-) \hspace{0.5cm} 7u + 4v= 2}$

$\displaystyle 20v = 5$

$\displaystyle \Rightarrow v = \frac{1}{4}$

$\displaystyle \text{Substituting } v = \frac{1}{4}$ in (i) we get

$\displaystyle 7u + 24( \frac{1}{4} ) = 7 \Rightarrow u = \frac{1}{7}$

Therefore

$\displaystyle \frac{1}{2x+3y} = u \Rightarrow 2x+3y = 7$ … … … … … (iii)

$\displaystyle \frac{1}{3x-2y} = v \Rightarrow 3x-2y = 4$ … … … … … (iv)

Now multiplying equation( (iv) by $\displaystyle 2$ and equation (iii) by $\displaystyle 3$ we get

$\displaystyle 6x+9y=21$ … … … … … (v)

$\displaystyle 6x-4y=8$ … … … … … (vi)

Subtracting (vi) from (v) we get

$\displaystyle 6x+9y=21$

$\displaystyle \underline{ (-) \hspace{0.5cm} 6x-4y=8}$

$\displaystyle 13y = 13$

$\displaystyle \Rightarrow y = 1$

$\displaystyle \text{Substituting } y =1$ in (v) we get

$\displaystyle 6x+9(1)=21 \Rightarrow x = 2$

$\displaystyle \text{Therefore } x =2, y = 1$ is the solution for the given system of equations.

$\displaystyle \\$

(iv) $\displaystyle 0.4x+0.3y=1.7 \text{ and } 0.7x-0.2y=0.8$

The given system of equation is:

$\displaystyle 0.4x+0.3y=1.7$ … … … … … (i)

$\displaystyle 0.7x-0.2y=0.8$ … … … … … (ii)

Now multiplying equation( (ii) by $\displaystyle 30$ and equation (i) by $\displaystyle 20$ we get

$\displaystyle 8x+6y=34$ … … … … … (iii)

$\displaystyle 21x-6y=24$ … … … … … (iv)

Adding (iv) and (iii) we get

$\displaystyle 8x+6y=34$

$\displaystyle \underline{ (+) \hspace{0.5cm} 21x-6y=24}$

$\displaystyle 29x = 58$

$\displaystyle \Rightarrow x = 2$

$\displaystyle \text{Substituting } x =2$ in (iii) we get

$\displaystyle 8(2)+6y=34 \Rightarrow y = 3$

$\displaystyle \text{Therefore } x =2, y = 3$ is the solution for the given system of equations.

$\displaystyle \\$

(v) $\displaystyle \frac{x}{2} + y = 0.8 \text{ and } \frac{7}{x+\frac{y}{2}} = 10$

The given system of equation is:

$\displaystyle \frac{x}{2} + y = 0.8 \Rightarrow x + 2y = 1.6$ … … … … … (i)

$\displaystyle \frac{7}{x+\frac{y}{2}} = 10 \Rightarrow 2x+y = 1.4$ … … … … … (ii)

Now multiplying equation( (i) by $\displaystyle 2$ and equation (ii) by $\displaystyle 1$ we get

$\displaystyle 2x+4y=3.2$ … … … … … (iii)

$\displaystyle 2x+y=1.4$ … … … … … (iv)

Subtracting (iv) and (iii) we get

$\displaystyle 2x+4y=3.2$

$\displaystyle \underline{ (-) \hspace{0.5cm} 2x+y=1.4}$

$\displaystyle 3y = 1.8$

$\displaystyle \Rightarrow y = 0.6$

$\displaystyle \text{Substituting } y = 0.6$ in (i) we get

$\displaystyle x = 1.6 - 2(0.6) = 0.4$

$\displaystyle \text{Therefore } x =0.4, y = 0.6$ is the solution for the given system of equations.

$\displaystyle \\$

(vi) $\displaystyle \frac{x}{3} + \frac{y}{4} = 11 \text{ and } \frac{5x}{6} - \frac{y}{3} = -7$

The given system of equation is:

$\displaystyle \frac{x}{3} + \frac{y}{4} = 11 \Rightarrow 4x+3y=132$ … … … … … (i)

$\displaystyle \frac{5x}{6} - \frac{y}{3} = -7 \Rightarrow 5x-2y=-42$ … … … … … (ii)

Now multiplying equation( (i) by $\displaystyle 2$ and equation (ii) by $\displaystyle 3$ we get

$\displaystyle 8x+6y = 264$ … … … … … (iii)

$\displaystyle 15x-6y = -126$ … … … … … (iv)

Adding (iv) and (iii) we get

$\displaystyle 8x+6y = 264$

$\displaystyle \underline{ (+) \hspace{0.5cm} 15x-6y = -126}$

$\displaystyle 23x = 138$

$\displaystyle \Rightarrow x = 6$

$\displaystyle \text{Substituting } x = 6$ in (i) we get

$\displaystyle 3y = 132 -4(6) -= 132 - 24 = 108 \Rightarrow y = 36$

$\displaystyle \text{Therefore } x =, y = 36$ is the solution for the given system of equations.

$\displaystyle \\$

(vii) $\displaystyle 3x- \frac{y+7}{11} + 2 = 10 \text{ and } 2y + \frac{x+11}{7} = 10$

The given system of equation is:

$\displaystyle 3x- \frac{y+7}{11} + 2 = 10 \Rightarrow 33x-y=95$ … … … … … (i)

$\displaystyle 2y + \frac{x+11}{7} = 10 \Rightarrow x+14y = 59$ … … … … … (ii)

Now multiplying equation( (i) by $\displaystyle 14$ and equation (ii) by $\displaystyle 1$ we get

$\displaystyle 462x-14y=1330$ … … … … … (iii)

$\displaystyle x+14y=59$ … … … … … (iv)

Adding (iv) and (iii) we get

$\displaystyle 462x-14y=1330$

$\displaystyle \underline{ (+) \hspace{0.5cm} x+14y=59}$

$\displaystyle 463 x = 1389$

$\displaystyle \Rightarrow x = 3$

$\displaystyle \text{Substituting } x = 3$ in (i) we get

$\displaystyle y = 33(3) - 95 = 99 - 95 = 4$

$\displaystyle \text{Therefore } x =3, y = 4$ is the solution for the given system of equations.

$\displaystyle \\$

(viii) $\displaystyle \frac{1}{2x} - \frac{1}{y} = -1 \text{ and } \frac{1}{x} + \frac{1}{2y} = 8$

$\displaystyle \text{Taking } \frac{1}{x} = u \text{ and } \frac{1}{y} = v \text{ , the given system of equations becomes }$

$\displaystyle u - 2v = -2$ … … … … … (i)

$\displaystyle 2u+v=16$ … … … … … (ii)

Now multiplying equation (i) by $\displaystyle 2$ and subtracting (ii) from equation (i) we get

$\displaystyle 2u-4v=-4$

$\displaystyle \underline{ (-) \hspace{0.5cm} 2u+v=16}$

$\displaystyle -5v=-20$

$\displaystyle \Rightarrow v = 4$

$\displaystyle \text{Substituting } v = 4$ in (ii) we get

$\displaystyle u = 2(4)-2 = 6$

$\displaystyle \text{Therefore } x = \frac{1}{6} \text{ and } y = \frac{1}{4}$

$\displaystyle \\$

(ix) $\displaystyle \frac{6}{x+y} = \frac{7}{x-y} + 3 \text{ and } \frac{1}{2(x+y)} = \frac{1}{3(x-y)}$

$\displaystyle \text{Taking } \frac{1}{x+y} = u \text{ and } \frac{1}{x-y} = v \text{ , the given system of equations becomes }$

$\displaystyle 6u-7v=3$ … … … … … (i)

$\displaystyle 3u -2v =0$ … … … … … (ii)

Now multiplying equation (ii) by $\displaystyle 2$ and subtracting it from equation (i) we get

$\displaystyle 6u-7v=3$

$\displaystyle \underline{ (-) \hspace{0.5cm} 3u -2v =0}$

$\displaystyle -3v= 3$

$\displaystyle \Rightarrow v = -1$

$\displaystyle \text{Substituting } v = -1$ in (ii) we get

$\displaystyle u = \frac{2}{3} (-1) = - \frac{2}{3}$

$\displaystyle \text{Therefore } x+y = - \frac{3}{2}$ … … … … … (iii)

$\displaystyle x-y = -1$ … … … … … (vi)

Adding (iii) and (iv) we get $\displaystyle 2x = - \frac{5}{2} \Rightarrow x = - \frac{5}{4}$

from (iv) $\displaystyle y = x+1 = - \frac{5}{4} + 1 = - \frac{1}{4}$

$\displaystyle \\$

(x) $\displaystyle \frac{5}{x+y} - \frac{2}{x-y} = -1 \text{ and } \frac{15}{x+y} + \frac{7}{x-y} = 10$

$\displaystyle \text{Taking } \frac{1}{x+y} = u \text{ and } \frac{1}{x-y} = v \text{ , the given system of equations becomes }$

$\displaystyle 5u-2v=-1$ … … … … … (i)

$\displaystyle 15u+7v=10$ … … … … … (ii)

Now multiplying equation (i) by $\displaystyle 3$ and subtracting (ii) from equation (i) we get

$\displaystyle 15u-6v=-3$

$\displaystyle \underline{ (-) \hspace{0.5cm} 15u+7v=10}$

$\displaystyle -13v=-13$

$\displaystyle \Rightarrow v = 1$

$\displaystyle \text{Substituting } v = 1$ in (ii) we get

$\displaystyle u = \frac{1}{5} (2-1) = \frac{1}{5}$

$\displaystyle \text{Therefore } x+y = 5$ … … … … … (iii)

$\displaystyle x-y = 1$ … … … … … (vi)

Adding (iii) and (iv) we get $\displaystyle 2x = 6 \Rightarrow x = 3$

from (iv) $\displaystyle y = x-1 = 3-1 = 2$

$\displaystyle \text{Therefore } x = 3, y = 2$

$\displaystyle \\$

(xi) $\displaystyle \frac{1}{2(x+2y)} + \frac{5}{3(3x-2y)} = -\frac{3}{2} \text{ and } \frac{5}{4(x+2y)} - \frac{3}{5(3x-2y)} = \frac{61}{60}$

$\displaystyle \text{Taking } \frac{1}{x+2y} = u \text{ and } \frac{1}{3x-2y} = v \text{ , the given system of equations becomes }$

$\displaystyle 3u+10v=-9$ … … … … … (i)

$\displaystyle 75u-36v=61$ … … … … … (ii)

Now multiplying equation (i) by $\displaystyle 25$ and subtracting (ii) from equation (i) we get

$\displaystyle 75u+250v=-225$

$\displaystyle \underline{ (-) \hspace{0.5cm} 75u-36v=61}$

$\displaystyle 286v=-286$

$\displaystyle \Rightarrow v = -1$

$\displaystyle \text{Substituting } v = -1$ in (ii) we get

$\displaystyle 3u = -9 -10(-1) = 1 \Rightarrow u = \frac{1}{3}$

$\displaystyle \text{Therefore } x+2y = 3$ … … … … … (iii)

$\displaystyle 3x-2y = -1$ … … … … … (vi)

Adding (iii) and (iv) we get $\displaystyle 4x = 2 \Rightarrow x = \frac{1}{2}$

from (iv) $\displaystyle y = \frac{1}{2} (3 - \frac{1}{2} ) = \frac{5}{4}$

$\displaystyle \text{Therefore } x = \frac{1}{2} , y = \frac{5}{4}$

$\displaystyle \\$

(xii) $\displaystyle \frac{2}{3x+2y} + \frac{3}{3x-2y} = \frac{17}{5} \text{ and } \frac{5}{3x+2y} + \frac{1}{3x-2y} = 2$

$\displaystyle \text{Taking } \frac{1}{3x+2y} = u \text{ and } \frac{1}{3x-2y} = v \text{ , the given system of equations becomes }$

$\displaystyle 10u+15v=17$ … … … … … (i)

$\displaystyle 5u+v=2$ … … … … … (ii)

Now multiplying equation (ii) by $\displaystyle 2$ and subtracting (ii) from equation (i) we get

$\displaystyle 10u+15v=17$

$\displaystyle \underline{ (-) \hspace{0.5cm} 10u+2v=4}$

$\displaystyle 13v = 13$

$\displaystyle \Rightarrow v = 1$

$\displaystyle \text{Substituting } v = 1$ in (ii) we get

$\displaystyle 5u = 2-1= 1 \Rightarrow u = \frac{1}{5}$

$\displaystyle \text{Therefore } 3x+2y = 5$ … … … … … (iii)

$\displaystyle 3x-2y = 1$ … … … … … (vi)

Adding (iii) and (iv) we get $\displaystyle 6x = 6 \Rightarrow x = 1$

from (iii) $\displaystyle 2y = 5-3(1) = 2 \Rightarrow y = 1$

$\displaystyle \text{Therefore } x = 1, y = 1$

$\displaystyle \\$

(xiii) $\displaystyle \frac{5}{x+1} - \frac{2}{y-1} = \frac{1}{2} \text{ and } \frac{10}{x+1} + \frac{2}{y-1} = \frac{5}{2}$

$\displaystyle \text{Taking } \frac{1}{x+1} = u \text{ and } \frac{1}{y-1} = v \text{ , the given system of equations becomes }$

$\displaystyle 10u-4v=1$ … … … … … (i)

$\displaystyle 20u+4v=5$ … … … … … (ii)

Adding (i) and (ii) we get

$\displaystyle 10u-4v=1$

$\displaystyle \underline{ (+) \hspace{0.5cm} 20u+4v=5}$

$\displaystyle 30u = 6$

$\displaystyle \Rightarrow u = \frac{1}{5}$

$\displaystyle \text{Substituting } u = \frac{1}{5}$ in (i) we get

$\displaystyle 4v = 10(\frac{1}{5})-1 = 2-1 = 1 \Rightarrow v = \frac{1}{4}$

$\displaystyle \text{Therefore } x+1 = 5 \Rightarrow x = 4$

$\displaystyle y-1 = 4 \Rightarrow y = 5$

$\displaystyle \text{Therefore } x = 4, y = 5$

$\displaystyle \\$

(xiv) $\displaystyle x-y+z=4$ , $\displaystyle x-2y-2z=9 \text{ and } 2x+y+3z=1$

Given system of equations is

$\displaystyle x-y+z=4$ … … … … … (i)

$\displaystyle x-2y-2z=9$ … … … … … (ii)

$\displaystyle 2x+y+3z=1$ … … … … … (iii)

From (i) $\displaystyle z = 4 - x +y$

Substituting this in (ii) and (iii) we get

$\displaystyle 3x-4y = 17$ … … … … … (iv)

$\displaystyle x-4y = 11$ … … … … … (v)

Subtracting (v) from (iv) we get

$\displaystyle 3x-4y = 17$

$\displaystyle \underline{ (-) \hspace{0.5cm} x-4y = 11}$

$\displaystyle x= 3$

From (v) $\displaystyle 4y = 3 - 11 = -8 \Rightarrow y = -2$

$\displaystyle \text{Therefore } z = 4 -3 + (-2) = -1$

$\displaystyle \text{Hence } x = 3, y = -2 \ and \ z = -1$

$\displaystyle \\$

(xv) $\displaystyle x-y+z=4$ , $\displaystyle x+y+z=2 \text{ and } 2x+y-3z=0$

Given system of equations is

$\displaystyle x-y+z=4$ … … … … … (i)

$\displaystyle x+y+z=2$ … … … … … (ii)

$\displaystyle 2x+y-3z=0$ … … … … … (iii)

From (i) $\displaystyle y = x+z-4$

Substituting this in (ii) and (iii) we get

$\displaystyle 2x+2z=6$ … … … … … (iv)

$\displaystyle 3x-2z=4$ … … … … … (v)

Adding (v) from (iv) we get

$\displaystyle 2x+2z=6$

$\displaystyle \underline{ (+) \hspace{0.5cm} 3x-2z=4}$

$\displaystyle 5x= 10$

$\displaystyle \text{Therefore } x = 2$

From (v) $\displaystyle 2z = 6-2(2) = 4 \Rightarrow z = 1$

$\displaystyle \text{Therefore } y = 2+1-4 = -1$

$\displaystyle \text{Hence } x = 2, y = -1 \ and \ z = 1$

$\displaystyle \\$

(xvi) $\displaystyle \frac{10}{x+y} + \frac{2}{x-y} = 4 \text{ and } \frac{15}{x+y} - \frac{9}{x-y} = -2$

$\displaystyle \text{Taking } \frac{1}{x+y} = u \text{ and } \frac{1}{x-y} = v \text{ , the given system of equations becomes }$

$\displaystyle 10u+2v=4$ … … … … … (i)

$\displaystyle 15u-9v=-2$ … … … … … (ii)

Multiplying (i) by $\displaystyle 3$ and (ii) by $\displaystyle 2$ we get

$\displaystyle 30u+6v=12$ … … … … … (iii)

$\displaystyle 30u-18v=-4$ … … … … … (iv)

subtracting (iv) from (iii)

$\displaystyle 30u+6v=12$

$\displaystyle \underline{ (-) \hspace{0.5cm} 30u-18v=-4}$

$\displaystyle 24v=16$

$\displaystyle \Rightarrow v = \frac{2}{3}$

$\displaystyle \text{Substituting } v = \frac{2}{3}$ in (ii) we get

$\displaystyle 10u = 4 - 2 ( \frac{2}{3} ) = \frac{8}{3} \Rightarrow u = \frac{4}{15}$

$\displaystyle \text{Therefore } x+y = \frac{15}{4}$ … … … … … (iii)

$\displaystyle x-y = \frac{3}{2}$ … … … … … (vi)

Adding (iii) and (iv) we get $\displaystyle 2x = \frac{21}{4} \Rightarrow x = \frac{21}{8}$

from (iv) $\displaystyle y = \frac{21}{8} - \frac{3}{2} = \frac{9}{8}$

$\displaystyle \text{Hence } x = \frac{21}{8} \text{ and } y = \frac{9}{8}$

$\displaystyle \\$

(xvii) $\displaystyle \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \text{ and } \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = - \frac{1}{8}$

$\displaystyle \text{Taking } \frac{1}{3x+y} = u \text{ and } \frac{1}{3x-y} = v \text{ , the given system of equations becomes }$

$\displaystyle 4u+4v=3$ … … … … … (i)

$\displaystyle 4u-4v=-1$ … … … … … (ii)

Adding (i) and (ii)

$\displaystyle 4u+4v=3$

$\displaystyle \underline{ (-) \hspace{0.5cm} 4u-4v=-1}$

$\displaystyle 8u=2$

$\displaystyle \Rightarrow u = \frac{1}{4}$

$\displaystyle \text{Substituting } u = \frac{1}{4}$ in (ii) we get

$\displaystyle v = \frac{3}{4} - \frac{1}{2} = \frac{1}{2}$

$\displaystyle \text{Therefore } 3x+y = 4$ … … … … … (iii)

$\displaystyle 3x-y = 2$ … … … … … (vi)

Adding (iii) and (iv) we get $\displaystyle 6x = 6 \Rightarrow x = 1$

from (iv) $\displaystyle y = 3(1) - 2 = 1$

$\displaystyle \\$

Question 3: Solve using the cross multiplication method.

(i) $\displaystyle ax+by= a-b \text{ and } bx-ay = a+b$

(ii) $\displaystyle x+y = (a+b) \text{ and } ax-by = a^2 - b^2$

(iii) $\displaystyle ax+by = 1 \text{ and } bx+ay = \frac{2ab}{a^2+b^2}$

(iv) $\displaystyle \frac{a}{x} - \frac{b}{y} = 0 \text{ and } \frac{ab^2}{x} + \frac{a^2b}{y} = a^2+b^2$ where $\displaystyle x, y \neq 0$

(v) $\displaystyle 3x+2y+25=0 \text{ and } 2x+y+10=0$

(vi) $\displaystyle ax+by = a^2 \text{ and } bx+ay=b^2$

(vii) $\displaystyle (a+2b)x+(2a-b)y=2 \text{ and } (a-2b)x+(2a+b)y=3$

(viii) $\displaystyle x(a-b+ \frac{ab}{a-b} )= y(a+b- \frac{ab}{a+b} ) \text{ and } x+y=2a^2$

(ix) $\displaystyle bx+cy = a+b \text{ and } ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}$

(x) $\displaystyle (a-b)x+(a+b)y = 2a^2 - 2b^2 \text{ and } (a+b)(x+y) = 4ab$

(xi) $\displaystyle \frac{a^2}{x} - \frac{b^2}{y} = 0 \text{ and } \frac{a^2b}{x} + \frac{b^2a}{y} = a+b$ where $\displaystyle x, y \neq 0$

Answers:

(i) $\displaystyle ax+by= a-b \text{ and } bx-ay = a+b$

The given system of equations may be written as

$\displaystyle ax+by -(a-b) = 0$

$\displaystyle bx-ay - (a+b)= 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ b ] \times [ -(a+b) ] - [-a] \times [ -(a-b) ]} = \frac{-y}{[ a ] \times [ -(a+b) ] - [b] \times [ -(a-b) ]} = \frac{1}{[a] \times [-a]-[b] \times [b]}$

$\displaystyle \Rightarrow \frac{x}{-ab-b^2-a^2+ab} = \frac{-y}{-a^2-ab+ab-b^2} = \frac{1}{-a^2-b^2}$

$\displaystyle \Rightarrow \frac{x}{-b^2-a^2} = \frac{-y}{-a^2-b^2} = \frac{1}{-(a^2+b^2)}$

$\displaystyle \Rightarrow \frac{x}{-(a^2+b^2)} = \frac{-y}{-(a^2+b^2)} = \frac{1}{-(a^2+b^2)}$

$\displaystyle \Rightarrow x = \frac{-(a^2+b^2)}{-(a^2+b^2)} = 1 \text{ and } y = - \frac{-(a^2+b^2)}{-(a^2+b^2)} = -1$

$\displaystyle \\$

(ii) $\displaystyle x+y = (a+b) \text{ and } ax-by = a^2 - b^2$

The given system of equations may be written as

$\displaystyle x+y - (a+b) = 0$

$\displaystyle ax-by -(a^2 - b^2)= 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ 1 ] \times [ -(a^2 - b^2) ] - [-b] \times [ - (a+b) ]} = \frac{-y}{[ 1 ] \times [ -(a^2 - b^2) ] - [a] \times [ - (a+b) ]} = \frac{1}{[1] \times [-b]-[1] \times [a]}$

$\displaystyle \Rightarrow \frac{x}{-a^2+b^2-ab-b^2} = \frac{-y}{-a^2+b^2+a^2+ab} = \frac{1}{-b-a}$

$\displaystyle \Rightarrow \frac{x}{-a(a+b)} = \frac{-y}{b(a+b)} = \frac{1}{-(a+b)}$

$\displaystyle \Rightarrow x = \frac{-a(a+b)}{-(a+b)} = a \text{ and } y = - \frac{b(a+b)}{-(a+b)} = b$

$\displaystyle \\$

(iii) $\displaystyle ax+by = 1 \text{ and } bx+ay = \frac{2ab}{a^2+b^2}$

The given system of equations may be written as

$\displaystyle ax+by - 1 = 0$

$\displaystyle bx+ay - \frac{2ab}{a^2+b^2} = 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ b ] \times [ -\frac{2ab}{a^2+b^2} ] - [a] \times [ -1 ]} = \frac{-y}{[ a ] \times [ -\frac{2ab}{a^2+b^2} ] - [b] \times [ - 1 ]} = \frac{1}{[a] \times [a]-[b] \times [b]}$

$\displaystyle \Rightarrow \frac{x}{-\frac{2ab}{a^2+b^2}+a} = \frac{-y}{-\frac{2ab}{a^2+b^2}+b} = \frac{1}{a^2-b^2}$

$\displaystyle \Rightarrow \Big( \frac{x}{\frac{-2ab^2+a^3+ab^2}{a^2+b^2}} \Big) = \Big( \frac{-y}{\frac{-2a^2b+b^3+a^2b}{a^2+b^2}} \Big) = \frac{1}{a^2-b^2}$

$\displaystyle \Rightarrow \Big( \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}} \Big) = \Big( \frac{-y}{\frac{b(b^2-a^2)}{a^2+b^2}} \Big) = \frac{1}{a^2-b^2}$

$\displaystyle \Rightarrow x = \frac{a(a^2-b^2)}{a^2+b^2} \times \frac{1}{a^2-b^2} = \frac{a}{a^2+b^2}$

$\displaystyle \Rightarrow y = - \frac{b(b^2-a^2)}{a^2+b^2} \times \frac{1}{a^2-b^2} = \frac{b}{a^2+b^2}$

$\displaystyle \\$

(iv) $\displaystyle \frac{a}{x} - \frac{b}{y} = 0 \text{ and } \frac{ab^2}{x} + \frac{a^2b}{y} = a^2+b^2$ where $\displaystyle x, y \neq 0$

Let $\displaystyle \frac{1}{x} =u \text{ and } \frac{1}{y} = v$

The given system of equations may be written as

$\displaystyle au-bv + 0 = 0$

$\displaystyle ab^2u + a^2bv - (a^2+b^2) = 0$

By Cross multiplication, we get

$\displaystyle \frac{u}{[ -b ] \times [ -(a^2+b^2) ] - [a^2b] \times [ 0 ]} = \frac{-v}{[ a ] \times [ -(a^2+b^2) ] - [ab^2] \times [ 0 ]} = \frac{1}{[a] \times [a^2b]-[ab^2] \times [-b]}$

$\displaystyle \Rightarrow \frac{u}{a^2+b^3} = \frac{-v}{-a^3-ab^2} = \frac{1}{a^3b+ab^3}$

$\displaystyle \Rightarrow u = \frac{b(a^2+b^2)}{ab(a^2+b^2)} = \frac{1}{a} \Rightarrow x = a$

$\displaystyle \Rightarrow -v = \frac{-a(a^2+b^2)}{ab(a^2+b^2)} = \frac{-1}{b} \Rightarrow y = b$

$\displaystyle \\$

(v) $\displaystyle 3x+2y+25=0 \text{ and } 2x+y+10=0$

The given system of equations is

$\displaystyle 3x+2y+25=0$

$\displaystyle 2x+y+10=0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ 2 ] \times [ 10 ] - [1] \times [ 25 ]} = \frac{-y}{[3] \times [ 10 ]- [ 2 ] \times [ 25 ]} = \frac{1}{[3] \times [1]-[2] \times [2]}$

$\displaystyle \Rightarrow \frac{x}{-5} = \frac{-y}{-20} = \frac{1}{-1}$

$\displaystyle \Rightarrow x = \frac{-5}{-1} = 5$

$\displaystyle \Rightarrow y = - ( \frac{-20}{-1} ) = -20$

$\displaystyle \\$

(vi) $\displaystyle ax+by = a^2 \text{ and } bx+ay=b^2$

The given system of equations may be written as

$\displaystyle ax+by - a^2=0$

$\displaystyle bx+ay-b^2=0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ b ] \times [ -b^2 ] - [a] \times [-a^2 ]} = \frac{-y}{[b] \times [ -a^2 ]- [ a ] \times [ -b^2 ]} = \frac{1}{[a] \times [a]-[b] \times [b]}$

$\displaystyle \Rightarrow \frac{x}{-b^3+a^3} = \frac{-y}{-ba^2+ab^2} = \frac{1}{a^2-b^2}$

$\displaystyle \Rightarrow x = \frac{a^3-b^3}{a^2-b^2} = \frac{(a-b)(a^2+ab+b^2)}{(a+b)(a-b)} = \frac{a^2+ab+b^2}{a+b}$

$\displaystyle \Rightarrow y = - \frac{-ba^2+ab^2}{a^2-b^2} = \frac{-ab(a-b)}{(a+b)(a-b)} = \frac{-ab}{a+b}$

$\displaystyle \\$

(vii) $\displaystyle (a+2b)x+(2a-b)y=2 \text{ and } (a-2b)x+(2a+b)y=3$

The given system of equations may be written as

$\displaystyle (a+2b)x+(2a-b)y - 2 = 0$

$\displaystyle (a-2b)x+(2a+b)y - 3 = 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ 2a-b ] \times [ -3 ] - [2a+b] \times [ -2 ]} = \frac{-y}{[ a+2b ] \times [ -3 ] - [a-2b] \times [ - 2 ]} = \frac{1}{[a+2b] \times [2a+b]-[a-2b] \times [2a-b]}$

$\displaystyle \Rightarrow \frac{x}{-6a+3b+4a+2b} = \frac{-y}{-3a-6b+2a-4b} = \frac{1}{2a^2+5ab+2b^2 - 2a^2 +5ab - 2b^2}$

$\displaystyle \Rightarrow \frac{x}{-2a+5b} = \frac{-y}{-a-10b} = \frac{1}{10ab}$

$\displaystyle \Rightarrow x = \frac{-2a+5b}{10ab}$

$\displaystyle \Rightarrow y = \frac{a+10b}{10ab}$

$\displaystyle \\$

(viii) $\displaystyle x(a-b+ \frac{ab}{a-b} )= y(a+b- \frac{ab}{a+b} ) \text{ and } x+y=2a^2$

We can simplify the equation first:

$\displaystyle x(a-b+ \frac{ab}{a-b} )= y(a+b- \frac{ab}{a+b} )$

$\displaystyle \Rightarrow x ( \frac{a^2+b^2-ab}{a-b} ) - y ( \frac{a^2+b^2 +ab}{a+b} ) +0 = 0$

$\displaystyle \Rightarrow x( \frac{a^3+b^3}{a^2-b^2} ) - y ( \frac{a^3-b^3}{a^2-b^2} ) +0 = 0$

The given system of equations may be written as

$\displaystyle \Rightarrow x( \frac{a^3+b^3}{a^2-b^2} ) - y ( \frac{a^3-b^3}{a^2-b^2} ) +0 = 0$

$\displaystyle x+y-2a^2 = 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ -\frac{a^3-b^3}{a^2-b^2} ] \times [-2a^2 ] - [0] \times [ 1 ]} = \frac{-y}{[ \frac{a^3+b^3}{a^2-b^2}] \times [ -2a^2 ] - [0] \times [ 1 ]} = \frac{1}{[\frac{a^3+b^3}{a^2-b^2}] \times [1]-[-\frac{a^3-b^3}{a^2-b^2}] \times [1]}$

$\displaystyle \Rightarrow \frac{x}{\frac{(a^3-b^3)(2a^2)}{a^2-b^2} } = \frac{-y}{\frac{(a^3+b^3)(-2a^2)}{a^2-b^2} } = \frac{1}{\frac{2a^3}{a^2-b^2} }$

$\displaystyle \Rightarrow \frac{x}{a^3-b^3} = \frac{y}{a^3+b^3} = \frac{1}{a}$

$\displaystyle \Rightarrow x = \frac{a^3-b^3}{a}$

$\displaystyle \Rightarrow y = \frac{a^3-b^3}{a}$

$\displaystyle \\$

(ix) $\displaystyle bx+cy = a+b \text{ and } ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}$

First simplify the equation:

$\displaystyle ax( \frac{1}{a-b} - \frac{1}{a+b} ) + cy ( \frac{1}{b-a} - \frac{1}{b+a} ) = \frac{2a}{a+b}$

$\displaystyle \Rightarrow ax ( \frac{a+b-a+b}{a^2-b^2} ) + cy ( \frac{b+a-b+a}{b^2-a^2} ) = \frac{2a}{a+b}$

$\displaystyle \Rightarrow \frac{2ab}{a^2-b^2} x - \frac{2ac}{a^2-b^2} y = \frac{2a}{a+b}$

$\displaystyle \Rightarrow \frac{bx}{a-b} - \frac{cy}{a-b} = 1$

The given system of equations may be written as

$\displaystyle \Rightarrow \frac{bx}{a-b} - \frac{cy}{a-b} - 1 = 0$

$\displaystyle bx+cy-(a+b) = 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ -\frac{c}{a-b} ] \times [-(a+b) ] - [-1] \times [ c ]} = \frac{-y}{[ \frac{b}{a-b}] \times [ -(a+b) ] - [-1] \times [ b ]} = \frac{1}{[\frac{b}{a-b}] \times [c]-[-\frac{c}{a-b}] \times [b]}$

$\displaystyle \Rightarrow \frac{x}{\frac{c(a+b)}{a-b}+c } = \frac{-y}{\frac{-b(a+b)}{a-b}+b } = \frac{1}{\frac{bc}{a-b} +\frac{bc}{a-b} }$

$\displaystyle \Rightarrow \frac{x}{c(a+b)+c(a-b) } = \frac{-y}{-b(a+b)+b(a-b) } = \frac{1}{2bc}$

$\displaystyle \Rightarrow \frac{x}{2ca} = \frac{y}{2b^2} = \frac{1}{2bc}$

$\displaystyle \Rightarrow x = \frac{2ca}{2bc} = \frac{a}{b}$

$\displaystyle \Rightarrow y = \frac{2b^2}{2bc} = \frac{b}{c}$

$\displaystyle \\$

(x) $\displaystyle (a-b)x+(a+b)y = 2a^2 - 2b^2 \text{ and } (a+b)(x+y) = 4ab$

First simplify the equations:

$\displaystyle (a-b)x+(a+b)y = 2a^2 - 2b^2$

$\displaystyle \Rightarrow x = \frac{a+b}{a-b} y - 2 \frac{(a-b)(a+b)}{(a-b)} = 0$

$\displaystyle \Rightarrow x + \frac{a+b}{a-b} y -2(a+b) = 0$

Also $\displaystyle x + y - \frac{4ab}{a+b} = 0$

By Cross multiplication, we get

$\displaystyle \frac{x}{[ \frac{a+b}{a-b} ] \times [-\frac{4ab}{a+b}] - [-2(a+b)] \times [ 1 ]} = \frac{-y}{[ -2(a+b)] \times [ 1 ] - [1] \times [ -\frac{4ab}{a+b}]} = \frac{1}{[1] \times [1]-[\frac{a+b}{a-b}] \times [1]}$

$\displaystyle \Rightarrow \frac{x}{\frac{-4ab}{a-b}+2(a+b) } = \frac{-y}{\frac{4ab}{a+b}-2(a+b) } = \frac{1}{1 -\frac{a+b}{a-b} }$

$\displaystyle x = \frac{\frac{-4ab}{a-b}+2(a+b) }{1 -\frac{a+b}{a-b}} = \frac{-4ab + 2(a^2-b^2)}{a - b - a - b} = \frac{2ab-a^2+b^2}{b}$

$\displaystyle y = \frac{-2(a+b) + \frac{4ab}{a+b}}{1 - \frac{a+b}{a-b}} = \frac{-2(a+b)^2+4ab}{a-b-a-b} \times \frac{a-b}{a+b}$

$\displaystyle = \frac{-2a^2-2b^2-4ab+4ab}{-2b} \times \frac{a-b}{a+b} = \frac{(a^2+b^2)(a-b)}{b(a+b)}$

$\displaystyle \\$

(xi) $\displaystyle \frac{a^2}{x} - \frac{b^2}{y} = 0 \text{ and } \frac{a^2b}{x} + \frac{b^2a}{y} = a+b$ where $\displaystyle x, y \neq 0$

Let $\displaystyle \frac{1}{x} = u \text{ and } \frac{1}{y} = v$

Therefore the given set of equations can be written as:

$\displaystyle a^2u-b^2v+0 = 0$

$\displaystyle a^2bu + b^2av-(a+b) = 0$

By Cross multiplication, we get

$\displaystyle \frac{u}{[-b^2 ] \times [-(a+b)] - [0] \times [ b^2a]} = \frac{-v}{[ a^2] \times [-(a+b) ] - [0] \times [ a^2b]} = \frac{1}{[a^2] \times [-(a+b)-[-b^2] \times [a^2b]}$

$\displaystyle \Rightarrow \frac{u}{b^2(a+b)} = \frac{-v}{-a^2(a+b) } = \frac{1}{a^3b^2+b^3a^2}$

$\displaystyle \text{Therefore } u = \frac{b^2(a+b)}{ab(a+b)} = \frac{b}{a} \Rightarrow x = \frac{a}{b}$

$\displaystyle v = \frac{a^2(a+b)}{ab(a+b)} = \frac{a}{b} \Rightarrow y = \frac{b}{a}$

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Class 9: Simultaneous Linear Equations – Exercise 7.1

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