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How do you solve the system of linear equations $y=-8x-3$ and $y=4x-3$?

Last updated date: 20th Jun 2024
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Hint: There are two unknowns $x$ and $y$ and also two equations to solve. We are applying the process of substitution and then the reduction. We take the value of the one variable and place that on another equation to solve the variables. We solve the equations equating the coefficients of one variable and omitting the variable. The other variable remains with the constants. Using the binary operation, we find the value of the other variable.

Complete step-by-step solution:
Let’s take two equations and two unknowns. $ax+by=m$ and $cx+dy=n$
If $\dfrac{a}{c}\ne \dfrac{b}{d}$, then we have only one solution for those equations.
For equations $y=-8x-3$ and $y=4x-3$, we have $\dfrac{-8}{4}\ne \dfrac{1}{1}$.
So, these equations have only one solution.
The given equations $y=-8x-3$ and $y=4x-3$ are linear equations of two variables.
We know that the number of equations has to be equal to the number of unknowns to solve them.
We have the value of one variable $y$ with respect to $x$ based on the equation $y=-8x-3$. We replace the value of $y$ in the second equation of $y=4x-3$ and get
  & y=4x-3 \\
 & \Rightarrow -8x-3=4x-3 \\
We get the equation of $x$ and solve
  & \Rightarrow -8x-3=4x-3 \\
 & \Rightarrow 12x=0 \\
 & \Rightarrow x=0 \\
Putting the value of $x$ we get $y=4x-3=-3$.
Therefore, the values are $x=0,y=-3$.

Note: Now we solve it through a reduction method.
We subtract these equations to get
  & y-y=\left( -8x-3 \right)-\left( 4x-3 \right) \\
 & \Rightarrow -8x-3=4x-3 \\
 & \Rightarrow 12x=0 \\
 & \Rightarrow x=0 \\
The value of $y$ is 0. Putting the value in the equation $y=4x-3$, we get $y=4x-3=-3$.
Therefore, the values are $x=0,y=-3$.