Question

p + 2q = −9 and 3p + 4q + 17 = 0 solve by Cross-Multiplication Method

Answer 1

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Step-by-step explanation:

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Answer 2

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

Given pair of equations are

$\rm :\longmapsto\:p + 2q = - 9$

and

$\rm :\longmapsto\:3p + 4q + 17 = 0$

can be rewritten as

$\rm :\longmapsto\:3p + 4q = - 17$

Now, we have two equations as

$\red{\rm :\longmapsto\:p + 2q = - 9}$

and

$\red{\rm :\longmapsto\:3p + 4q = - 17}$

Now, using Cross multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf 2 & \sf - 9 & \sf 1 & \sf 2\\ \\ \sf 4 & \sf - 17 & \sf 3 & \sf 4\\ \end{array}} \\ \end{gathered}$

Now,

$\rm :\longmapsto\:\dfrac{p}{ - 34 - ( - 36)} = \dfrac{q}{ - 27 - ( - 17)} = \dfrac{ - 1}{4 - 6}$

$\rm :\longmapsto\:\dfrac{p}{ - 34 + 36} = \dfrac{q}{ - 27 + 17} = \dfrac{ - 1}{ - 2}$

$\rm :\longmapsto\:\dfrac{p}{2} = \dfrac{q}{ - 10} = \dfrac{ 1}{ 2}$

On Comparing first and third member, we get

$\rm :\longmapsto\:\dfrac{p}{2} = \dfrac{ 1}{ 2}$

$\bf\implies \:p = 1$

Now, Comparing second and third member, we get

$\rm :\longmapsto\: \dfrac{q}{ - 10} = \dfrac{ 1}{ 2}$

$\bf\implies \:q = - 5$

VERIFICATION :-

Consider first equation,

$\rm :\longmapsto\:p + 2q = - 9$

On substituting the values of p and q, we get

$\rm :\longmapsto\:1 + 2( - 5) = - 9$

$\rm :\longmapsto\:1 - 10 = - 9$

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

Hence, Verified

Consider second equation,

$\rm :\longmapsto\:3p + 4q = - 17$

On substituting the values of p and q, we get

$\rm :\longmapsto\:3(1) + 4( - 5) = - 17$

$\rm :\longmapsto\:3 - 20 = - 17$

$\rm :\longmapsto\: - 17 = - 17$

Hence, Verified