Question

How do you solve $3x+4y=27$ and $-8x+y=33$ using substitution?

Answer 1

Hint: We have been given two linear equations in two 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 substitution method, we shall substitute the value of variable-y from the second equation to the first equation. After obtaining the value of x-variable, we shall substitute this value in the second equation to calculate the value of y-variable.

Complete step by step answer:
Given that
$3x+4y=27$ ……………… equation (1)
And $-8x+y=33$ ……………… equation (2)
In equation (2), we shall transpose the term with the x-variable to the right hand side and obtain this equation in terms of y-variable which will be further substituted into equation (1).
$\Rightarrow y=33+8x$
Substituting this value of y-variable in equation (1), we get
$\Rightarrow 3x+4(33+8x)=27$
Opening the brackets on the left hand side and multiplying each term by 4, we get
$\Rightarrow 3x+132+32x=27$
Here, we shall transpose the constant term, 132 to the right hand side and subtract it from 27.
$\Rightarrow 3x+32x=27-132$
$\Rightarrow 35x=-105$
Dividing both sides by 35 to obtain the x-coordinate of the point of intersection, we get
$\Rightarrow x=-{\displaystyle \frac{105}{35}}$
$\Rightarrow x=-3$
Now, we shall substitute the value of x-variable in equation (2) to obtain the value of y-coordinate.
$\Rightarrow -8(-3)+y=33$
$\Rightarrow 24+y=33$
Transposing the constant term, 24 to the right hand side, we get
$\Rightarrow y=33-24$
$\Rightarrow y=9$

Therefore, the solution or the point of intersection of $3x+4y=27$ and $-8x+y=33$ is $(-3,9)$.

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 (3,0) instead of (-3,0).