Question

Solve using cross multiplication.
4x-7y+28=0
7x-5y-9=0
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Answer 1

Answer:

4x-7y+28=0

7x-5y-9=0

Step-by-step explanation:

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Answer 2

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

➢Given pair of linear equations are

$\rm :\longmapsto\:4x - 7y + 28 = 0$

and

$\rm :\longmapsto\:7x - 5y - 9 = 0$

➢can be rewritten as

$\rm :\longmapsto\:4x - 7y = - 28$

and

$\rm :\longmapsto\:7x - 5y = 9$

➢Using Cross Multiplication method, we get

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf - 7 & \sf - 28 & \sf 4 & \sf - 7\\ \\ \sf - 5 & \sf 9 & \sf 7 & \sf - 5\\ \end{array}} \\ \end{gathered}$

So,

$\rm :\longmapsto\:\dfrac{x}{ - 63 - 140} = \dfrac{y}{ - 196 - 36} = \dfrac{ - 1}{ - 20 + 49}$

$\rm :\longmapsto\:\dfrac{x}{ - 203} = \dfrac{y}{ - 232} = \dfrac{ - 1}{29}$

$\rm :\longmapsto\:\dfrac{x}{ - 203} = \dfrac{y}{ - 232} = \dfrac{1}{ - 29}$

➢On multiply by (- 1), we get

$\rm :\longmapsto\:\dfrac{x}{203} = \dfrac{y}{ 232} = \dfrac{1}{ 29}$

➢On taking first and third member, we get

$\rm :\longmapsto\:\dfrac{x}{203} = \dfrac{1}{ 29}$

$\rm :\longmapsto\:x = \dfrac{203}{29}$

$\bf\implies \:x = 7$

➢Now, on taking second and third member, we get

$\rm :\longmapsto\: \dfrac{y}{ 232} = \dfrac{1}{ 29}$

$\rm :\longmapsto\:y = \dfrac{232}{29}$

$\bf\implies \:y = 8$

Verification :-

➢ Consider first equation,

$\rm :\longmapsto\:4x - 7y + 28 = 0$

➢ On substituting the values of x and y, we get

$\rm :\longmapsto\:4 \times 7 - 7 \times 8 + 28 = 0$

$\rm :\longmapsto\:28 - 56 + 28 = 0$

$\rm :\longmapsto\:0 = 0$

Hence, Verified

➢Consider second equation,

$\rm :\longmapsto\:7x - 5y - 9 = 0$

➢ On substituting the values of x and y, we get

$\rm :\longmapsto\:7 \times 7 - 5 \times 8 - 9 = 0$

$\rm :\longmapsto\:49 - 40 - 9 = 0$

$\rm :\longmapsto\:0 = 0$

Hence, Verified