Question

Solve the following:
$5x-4y+8=0$
$7x+6y-9=0$
A. $x=\dfrac{6}{29}$ and $y=-\dfrac{101}{58}$
B. $x=-\dfrac{6}{29}$ and $y=\dfrac{101}{58}$
C. $x=-\dfrac{8}{29}$ and $y=\dfrac{101}{58}$
D. None of these

Answer 1

Hint: We have been given two equations and we have to solve them. We will use the elimination method to solve the equations given. Firstly we will multiply the two equations by some numbers such that the coefficient of any one of the variables becomes the same. Then we will add the two equations and form a new equation with only one variable in it and solve it. Finally we will put the value obtained in any one of the original equations and get our desired answer.

Complete step by step answer:
The two equations are given as follows,
$5x-4y+8=0$….$\left( 1 \right)$
$7x+6y-9=0$……$\left( 2 \right)$
Using the elimination method we will solve the above equation.
We will make the coefficient of variable \[y\] the same by multiplying equation (1) by $3$ and equation (2) by $2$ as the smallest common multiple of $4$ and $6$ is $12$ .
So multiplying equation (1) by $3$ we get,
$\Rightarrow \left( 5x-4y+8=0 \right)\times 3$
$\Rightarrow 15x-12y+24=0$……$\left( 3 \right)$
Multiplying equation (2) by $2$ we get,
$\Rightarrow \left( 7x+6y-9=0 \right)\times 2$
$\Rightarrow 14x+12y-18=0$….$\left( 4 \right)$
On adding equation (3) and (4) we get,
$\begin{align}
  & 15x-12y+24=0 \\
 & \underline{+14x+12y-18=0} \\
 & 29x+0+6=0 \\
\end{align}$
So we get the new equation as,
$\Rightarrow 29x+6=0$
$\Rightarrow 29x=-6$
So we get,
$\Rightarrow x=-\dfrac{6}{29}$….$\left( 5 \right)$
On substituting the above value in equation (1) we get,
$\Rightarrow \left( 5\times -\dfrac{6}{29} \right)-4y+8=0$
$\Rightarrow -\dfrac{30}{29}-4y+8=0$
Taking constant term on right side we get,
$\Rightarrow -4y=\dfrac{30}{29}-8$
Taking LCM we get,
$\Rightarrow -4y=\dfrac{30-\left( 8\times 29 \right)}{29}$
$\Rightarrow -4y=\dfrac{30-232}{29}$
So we get,
$\Rightarrow -4y=\dfrac{-202}{29}$
$\Rightarrow y=\dfrac{-202}{-4\times 29}$
So,
$\Rightarrow y=\dfrac{101}{58}$…$\left( 6 \right)$
So from equation (5) and (6) we get $x=-\dfrac{6}{29}$ and $y=\dfrac{101}{58}$

So, the correct answer is “Option B”.

Note:
Elimination method is such that we add or subtract the two equations given in order to get a new equation with only one variable. This method is useful if the equation coefficient is not too big and we can easily figure out what number can be multiplied so that the coefficient of one variable becomes the same in both equations.
 If we are given two linear equations then we can solve them in many ways like using the substitution method and matrix method.