How do you solve a+3b=7 and 2a=b-7?

May 18, 2018

See a solution process below:

Explanation:

Step 1) Solve the first equation for $a$:

$a + 3 b = 7$

$a + 3 b - \textcolor{red}{3 b} = 7 - \textcolor{red}{3 b}$

$a + 0 = 7 - 3 b$

$a = 7 - 3 b$

Step 2) Substitute $\left(7 - 3 b\right)$ for $a$ in the second equation and solve for $b$:

$2 a = b - 7$ becomes:

$2 \left(7 - 3 b\right) = b - 7$

$\left(2 \cdot 7\right) - \left(2 \cdot 3 b\right) = b - 7$

$14 - 6 b = b - 7$

$14 + \textcolor{b l u e}{7} - 6 b + \textcolor{red}{6 b} = b + \textcolor{red}{6 b} - 7 + \textcolor{b l u e}{7}$

$21 - 0 = 1 b + \textcolor{red}{6 b} - 0$

$21 = \left(1 + \textcolor{red}{6}\right) b$

$21 = 7 b$

$\frac{21}{\textcolor{red}{7}} = \frac{7 b}{\textcolor{red}{7}}$

$3 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} b}{\cancel{\textcolor{red}{7}}}$

$3 = b$

$b = 3$

Step 3) Substitute $3$ for $b$ in the solution to the first equation at the end of Step 1 and calculate $a$:

$a = 7 - 3 b$ becomes:

$a = 7 - \left(3 \cdot 3\right)$

$a = 7 - 9$

$a = - 2$

The Solution Is:

$a = - 2$ and $b = 3$