Question

Solve by cross multiplication: 2x − 3y = 9; 4x + 9y + 2 = 0.

Answer 1

Answer:

(x, y) = ( 5/2, -4/3)

Step-by-step explanation:

Multiply equation (i) by 3 and add to equation (ii)

6x − 9y = 27     ... eq (i) x 3

4x + 9y = -2      ... eq (ii)

10x = 25

⇒ x = 25/10 = 5/2

∴ y = (-2 - 4x)/ 9 = (-2 -10 )/9 = -12/9 = -4/3

(x, y) = ( 5/2, -4/3)

Answer 2

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:2x - 3y = 9$

and

$\rm :\longmapsto\:4x + 9y + 2 = 0$

can be rewritten as

$\rm :\longmapsto\:4x + 9y = - \: 2$

Now, Using Cross Multiplication method, we get

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf - 3 & \sf 9 & \sf 2 & \sf - 3\\ \\ \sf 9 & \sf - 2 & \sf 4 & \sf 9\\ \end{array}} \\ \end{gathered}$

So, we have now

$\rm :\longmapsto\:\dfrac{x}{6 - 81} = \dfrac{y}{36 - ( - 4)} = \dfrac{ - 1}{18 - ( - 12)}$

$\rm :\longmapsto\:\dfrac{x}{ - 75} = \dfrac{y}{36 + 4} = \dfrac{ - 1}{18 + 12}$

$\rm :\longmapsto\:\dfrac{x}{ - 75} = \dfrac{y}{40} = \dfrac{ - 1}{30}$

Taking first and third member, we get

$\rm :\longmapsto\:\dfrac{x}{ - 75} = \dfrac{ - 1}{30}$

$\rm :\longmapsto\:{x} = \dfrac{75}{30}$

$\bf :\longmapsto\:{x} = \dfrac{5}{2}$

Taking second and third member, we get

$\rm :\longmapsto\: \dfrac{y}{40} = \dfrac{ - 1}{30}$

$\rm :\longmapsto\:{y} = \dfrac{ - 40}{30}$

$\bf :\longmapsto\:{y} = - \: \dfrac{ 4}{3}$

Verification :-

Consider,

$\rm :\longmapsto\:2x - 3y = 9$

On substituting the values of x and y, we get

$\rm :\longmapsto\:2 \times \dfrac{5}{2} - 3 \times \dfrac{ - 4}{3} = 9$

$\rm :\longmapsto\:5 - ( - 4) = 9$

$\rm :\longmapsto\:5 + 4 = 9$

$\rm :\longmapsto\:9 = 9$

Hence, Verified