Question

How do you solve by substitution $5x - 6y = 6$ and $5x + y = 2$?

Answer 1

Here we will proceed by taking one of the equations from the pair and convert it into a third equation. Then we will substitute the newly formed equation into another equation from the given pair of equations, it will give the value of one variable. After that substitute the value in the third equation to get the value of the second variable. Thus, we will get the required values of the equation.

Complete step-by-step answer:
Linear pairs of equations are equations that can be expressed as $ax + by + c = 0$ where a, b and c are real numbers and both a, b are non-zero.
In this question, two equations are-
$\Rightarrow 5x - 6y = 6$ ….. (1)
$\Rightarrow 5x + y = 2$ ….. (2)
Firstly, we will take equation (2) and convert it,
$\Rightarrow 5x + y = 2$
Subtract $5x$ from both sides of the equation,
$\Rightarrow 5x + y - 5x = 2 - 5x$
Simplify the term,
$\Rightarrow y = 2 - 5x$ ….. (3)
Now we will put the value of y from equation (3) in equation 1,
$\Rightarrow 5x - 6\left( {2 - 5x} \right) = 6$
Simplify the terms,
$\Rightarrow 5x - 12 + 30x = 6$
Move constant part on the right side,
$\Rightarrow 35x = 18$
Divide both sides by 35,
$\Rightarrow x = \dfrac{{18}}{{35}}$
Here we will substitute the value of x in equation (3) to get the value of y,
$\Rightarrow y = 2 - 5 \times \dfrac{{18}}{{35}}$
Cancel out the common factors,
$\Rightarrow y = 2 - \dfrac{{18}}{7}$
Take LCM on the right side,
$\Rightarrow y = \dfrac{{14 - 18}}{7}$
Simplify the term,
$\Rightarrow y = - \dfrac{4}{7}$

Hence values of x and y are $\dfrac{{18}}{{35}}$ and $- \dfrac{4}{7}$ respectively.

Note:
Whenever we face such types of problems the key concept is to use various methods of variable evaluation either by elimination or by substitution method. These methods will help in getting the right track to evaluate these equations involving two variables and reach the right solution.