Question

How do you solve the system $ y=2x+6 $ and $ 2x-y=2 $ using substitution?

Answer 1

Hint: There are two unknowns and two equations to solve. We solve the problem through substitution where we take one value of $ x $ from one equation and place it in the other equation. Then we verify the result by solving 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 answer:
The given equations $ y=2x+6 $ and $ 2x-y=2 $ 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 can also find the value of one variable y with respect to x based on the equation
 $ y=2x+6 $ . We replace the value of y for the substitution process in the second equation of
 $ 2x-y=2 $ and get
\[\begin{align}
  & 2x-y=2 \\
 & \Rightarrow 2x-\left( 2x+6 \right)=2 \\
 & \Rightarrow -6=2 \\
\end{align}\]
We get the equation without any variable to solve. The equations are not solvable.
The equations $ y=2x+6 $ and $ 2x-y=2 $ can be changed to $ -2x+y=6 $ and $ 2x-y=2 $ .
Now, for $ -2x+y=6 $ and $ 2x-y=2 $ . We take the coefficients and place them in the formula. The ratios of the respective coefficients are $ \dfrac{-2}{2},\dfrac{1}{-1},\dfrac{6}{2} $ .
We have $ \dfrac{-2}{2}=-1,\dfrac{1}{-1}=-1,\dfrac{6}{2}=3 $ .
The relation is $ \dfrac{-2}{2}=\dfrac{1}{-1}\ne \dfrac{6}{2} $ .
So, we have no solution to those equations.

Note:
 The coefficients of the given equation define the number of solutions for the given equations.
Let’s take two equations two unknowns. $ ax+by=m $ and $ cx+dy=n $
In the equations the terms a, b, c, d, m, n all are constants.
We take the ratios of the coefficients of same variables. The ratios are $ \dfrac{a}{c},\dfrac{b}{d},\dfrac{m}{n} $ .
The relation among these ratios determines the equations.
If $ \dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n} $ , then we have no solution for those equations.