Question

How do you solve $y=6x-4$ and $3x-2y=5$ using substitution?

Answer 1

Hint: We have two linear equations in two variables. We have to use the method of substitution to solve this system of linear equations. We will substitute one variable in one equation with its expression derived from the second equation. Then we will obtain a linear equation in one variable. Solving this equation, we will get the value of one variable. Substituting this value in one equation, we will obtain the value of the second variable.

Complete step-by-step answer:
We have a system of two linear equations in two variables. The given equations are the following,
$y=6x-4....(i)$
$3x-2y=5....(ii)$
We have to use the method of substitution to solve this system of linear equations. Let us substitute the value of $y$ from equation $(i)$ in equation $(ii)$. So, we have the following,
$3x-2\left( 6x-4 \right)=5$
Using the distributive property, we can simplify the above equation in the following manner,
$\begin{align}
  & 3x-2\times 6x+2\times 4=5 \\
 & \therefore 3x-12x+8=5 \\
\end{align}$
Solving the above equation for the variable $x$, we have
$\begin{align}
  & -9x=-3 \\
 & \therefore x=\dfrac{1}{3} \\
\end{align}$
Next, we will substitute this value of the variable $x$ in equation $(i)$ to obtain the value of the variable $y$. Substituting $x=\dfrac{1}{3}$ in equation $(i)$, we get the following,
$\begin{align}
  & y=6\left( \dfrac{1}{3} \right)-4 \\
 & \Rightarrow y=2-4 \\
 & \therefore y=-2 \\
\end{align}$
Hence, we get $x=\dfrac{1}{3}$ and $y=-2$ as the solution of the given system of linear equations.

Note: There are other methods of solving a system of linear equations. These methods are Gauss elimination method and graphing method. In the graphing method, we draw the lines represented by the two equations and the intersection of these lines is the solution. In the Gauss elimination method, we eliminate one variable and obtain the value of the other variable. Then we substitute this value in one equation to obtain the value of the first variable.