Question

Solve the following system of linear equations by substitution and determine whether the system has one solution, no solution, of allinfinite number of solutions. If the system has one solution, find the solution.4x + 2y = -2010x + 5y = -50

Answer 1

`\begin{gathered} 4x+2y=-20 \\ 10x+5y=-50 \end{gathered}`

To solve the above system of equation, let's use substitution method. Here are the steps:

1. Equate second equation to y = .

`\begin{gathered} 10x+5y=-50 \\ 5y=-10x-50 \\ \text{Divide both sides by 5.} \\ (5y)/(5)=-(10x)/(5)-(50)/(5) \\ y=-2x-10 \end{gathered}`

2. Use this value of "y" and plug it in to the first equation.

`\begin{gathered} 4x+2y=-20 \\ 4x+2(-2x-10)=-20 \end{gathered}`

Then, solve for x.

`\begin{gathered} 4x-4x-20=-20 \\ \text{Add 20 on both sides.} \\ 4x-4x-20+20=-20+20 \\ 0=0 \end{gathered}`

It seems like both equations are just equal to each other. With this, the system has infinite number of solutions.