Question

How do you solve the system of equations: $y = 3x - 10$ and $x = 12 - 4y$?

Answer 1

Hint: In this question, we are given two equations and we have been asked to solve them. You can use any method out of the three methods – elimination, substitution or cross multiplication. I will solve this question using two methods: elimination and substitution.
Substitution method: The two equations given are in terms of $x$ and $y$. Using one equation, find the value of one variable. Then, put this in the other equation such that the equation is in one variable only. Then, using this equation, find the value of that variable. Put this in any of the original equations and find the value of the other variable.

Complete step-by-step solution:
Let us solve by substitution method first. We have two equations –
$ \Rightarrow y = 3x - 10$ ……….…. (1)
$ \Rightarrow x = 12 - 4y$ ………..…. (2)
Put equation (1) in equation (2),
$ \Rightarrow x = 12 - 4\left( {3x - 10} \right)$
Now, we will simplify the equation to find the value of $x$.
$ \Rightarrow x = 12 - 12x + 40$
Taking variables as RHS and remaining as LHS we get,
$ \Rightarrow x + 12x = 12 + 40 = 52$
On adding we get
$ \Rightarrow 13x = 52$
Let us divide 13 on both sides we get
$ \Rightarrow x = \dfrac{{52}}{{13}} = 4$
Putting this in equation (1),
$ \Rightarrow y = 12 - 10 = 2$

Hence, $x = 4$ and $y = 2$.

Note: Here we have an alternative method as follows:
Elimination method: In this method, we eliminate one variable and find the value of the other. Choose any one variable. Make their coefficient same by multiplying or dividing a constant in the entire equation. Once it is done, subtract the equations. This will eliminate that variable which you chose to make the same. Find the value of the other variable. Put this in any of the original equations and find the value of the other variable.
Method 2: Now, we will solve the same question using the elimination method.
$ \Rightarrow y = 3x - 10$
Shifting the variables,
$ \Rightarrow 3x - y = 10$ …………... (1)
$ x = 12 - 4y$
Shifting the variables,
$ \Rightarrow x + 4y = 12$ ……………... (2)
Now, if we look at the equation (1), $x$ has a variable $3$. So, we will multiply equation (2) by 3 so as to make the coefficient same.
Multiplying equation (2) by 3,
$ \Rightarrow 3x + 12y = 36$ ……………... (3)
Now, we will subtract equation (1) from equation (3),
$ \Rightarrow 3x + 12y - \left( {3x - y} \right) = 36 - 10$
Solving the equation,
$ \Rightarrow 3x + 12y - 3x + y = 26$
On cancel the same term and adding the remaining term we get
$ \Rightarrow 13y = 26$
On dividing 13 on both sides we get,
$ \Rightarrow y = \dfrac{{26}}{{13}} = 2$
Putting this in equation (1),
$ \Rightarrow 3x - 2 = 10$
On adding 2 on both sides we get
$ \Rightarrow 3x = 12$
Let us divide 3 on both sides we get,
$ \Rightarrow x = \dfrac{{12}}{3} = 4$
Hence, $x = 4$ and $y = 2$.