Question

How do you solve $3x-2y=-21$ and $2x+5y=5$ using elimination?

Answer 1

Hint: We have been given two linear equations in 2-variables, variable-x and variable-y which must be solved simultaneously to find the point of intersection of the two given lines. In order to find this point of intersection by the elimination method, we shall multiply equation 1 by 5 and equation 2 by 2 to make the y-term of both the equations equal. Then while adding the modified equations, we shall eliminate the y-term and find the x-coordinate.

Complete step by step solution:
Given that
$3x-2y=-21$ ……………… equation (1)
And $2x+5y=5$ ……………… equation (2)
In equation (1), we shall multiply both sides by 5 and in equation (2), we shall multiply both sides by 2.
In equation (1):
$\Rightarrow 5\left( 3x-2y \right)=5\left( -21 \right)$
$\Rightarrow 15x-10y=-105$ ………………. Equation (3)
In equation (2):
$\Rightarrow 2\left( 2x+5y \right)=2\left( 5 \right)$
\[\Rightarrow 4x+10y=10\] ………………. Equation (4)
Now, we shall add equations (3) and (4), to eliminate the term of y-variable.
$\begin{align}
  & \text{ }15x-10y=-105 \\
 & +4x+10y\text{ }=10 \\
 & \text{ }\overline{19x+0y\text{ }=-95} \\
\end{align}$
$\Rightarrow 19x=-95$
Dividing both sides by 19 to determine the x-coordinate, we get
$\Rightarrow x=-\dfrac{95}{19}$
$\Rightarrow x=-5$
Substituting this value of x-variable in equation (2), we get
$\Rightarrow 2\left( -5 \right)+5y=5$
$\Rightarrow -10+5y=5$
Here, we shall transpose the constant term, -10 to the right hand side and add it to 5.
$\Rightarrow 5y=15$
Dividing both sides by 5 to obtain the y-coordinate of the point of intersection, we get
$\Rightarrow y=\dfrac{15}{5}$
$\Rightarrow y=3$
Therefore, the solution or the point of intersection of $3x-2y=-21$ and $2x+5y=5$ is $\left( -5,3 \right)$.

Note: Another method of finding the solution or the point of intersection of the given linear equations in two variables was by sketching the graph of both the straight-lines on the same cartesian plane. However, we must take care while marking the points on the graph. The possible mistake which can be made while sketching the graph would be marking (5,0) instead of (-5,0).