Question

Solve the following system of linear equations by substitution and determine whether the system has one solution, no solution, or aninfinite number of solutions. If the system has one solution, find the solution.4x + 4y = 42x+y=-7

Answer 1

`\begin{gathered} 4x+4y=4 \\ 2x+y=-7 \end{gathered}`

To solve the above system of equation, we can use substitution method. What we need to first is equate either of the two equations into y = . For this solution, we will equate the second equation.

`\begin{gathered} 2x+y=-7 \\ y=-2x-7 \end{gathered}`

Hence, for the second equation, y = -2x - 7. Let's use this value of "y" and plug it in to the first equation.

`\begin{gathered} 4x+4y=4 \\ \text{Plug in y = -2x - 7} \\ 4x+4(-2x-7)=4 \end{gathered}`

Then, solve for x.

`\begin{gathered} 4x-8x-28=4 \\ \text{Add 28 on both sides of the equation.} \\ 4x-8x-28+28=4+28 \\ -4x=32 \\ \text{Divide -4 on both sides of the equation.} \\ x=-8 \end{gathered}`

The value of x = -8.

To solve for y, we can use the second equation and plug in x = -8.

`\begin{gathered} y=-2x-7 \\ y=-2(-8)-7 \\ y=16-7 \\ y=9 \end{gathered}`

The value of y = 9.

Therefore, this system of equation has one solution and that is (-8, 9).