Question

Solve the following system of linear equations by substitution. (If there is no solution, enter NO SOLUTION.) 3a + 4b = 1a − 9b = 21

Answer 1

`\begin{gathered} 3a+4b=1 \\ a-9b=21 \end{gathered}`

To solve by substitution:

1. Solve a in the second equation:

`a=21+9b`

2. Substitute the a in the first equation by the value you get in the previous step:

`3(21+9b)+4b=1`

3. Solve b:

`\begin{gathered} 63+27b+4b=1 \\ 63+31b=1 \\ 31b=1-63 \\ 31b=-62 \\ b=(-62)/(31) \\ \\ b=-2 \end{gathered}`

4. Use the value of b to solve a:

`\begin{gathered} a=21+9b \\ a=21+9(-2) \\ a=21-18 \\ a=3 \end{gathered}`Then, the solution for the given system is:a=3b=-2