How do you solve the following system: 3x + 8y = -2, -2x + 5y = 20?

Oct 2, 2017

See a solution process below:

Explanation:

Step 1: Solve each equation for $6 x$:

• Equation 1: $3 x + 8 y = - 2$

$3 x + 8 y - \textcolor{red}{8 y} = - 2 - \textcolor{red}{8 y}$

$3 x + 0 = - 2 - 8 y$

$3 x = - 2 - 8 y$

$\textcolor{red}{2} \cdot 3 x = \textcolor{red}{2} \left(- 2 - 8 y\right)$

$6 x = \left(\textcolor{red}{2} \cdot - 2\right) - \left(\textcolor{red}{2} \cdot 8 y\right)$

$6 x = - 4 - 16 y$

• Equation 2: $- 2 x + 5 y = 20$

$- 2 x + 5 y - \textcolor{red}{5 y} = 20 - \textcolor{red}{5 y}$

$- 2 x + 0 = 20 - 5 y$

$- 2 x = 20 - 5 y$

$\textcolor{red}{- 3} \cdot - 2 x = \textcolor{red}{- 3} \left(20 - 5 y\right)$

$6 x = \left(\textcolor{red}{- 3} \cdot 20\right) - \left(\textcolor{red}{- 3} \cdot 5 y\right)$

$6 x = - 60 - \left(- 15 y\right)$

$6 x = - 60 + 15 y$

Step 2: Now with the left side of both equations equal we can equate the right side of the equations and solve for $y$:

$- 4 - 16 y = - 60 + 15 y$

$\textcolor{b l u e}{60} - 4 - 16 y + \textcolor{red}{16 y} = \textcolor{b l u e}{60} - 60 + 15 y + \textcolor{red}{16 y}$

$56 - 0 = 0 + \left(15 + \textcolor{red}{16}\right) y$

$56 = 31 y$

$\frac{56}{\textcolor{red}{31}} = \frac{31 y}{\textcolor{red}{31}}$

$\frac{56}{31} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{31}}} y}{\cancel{\textcolor{red}{31}}}$

$\frac{56}{31} = y$

$y = \frac{56}{31}$

Step 3: Substitute $\frac{56}{31}$ for $y$ in the solution to either equation in Step 1 and solve for $x$:

$6 x = - 60 + 15 y$ becomes:

$6 x = - 60 + \left(15 \cdot \frac{56}{31}\right)$

$6 x = \left(\frac{31}{31} \cdot - 60\right) + \frac{840}{31}$

$6 x = - \frac{1860}{31} + \frac{840}{31}$

$6 x = - \frac{1020}{31}$

$\textcolor{red}{\frac{1}{6}} \cdot 6 x = \textcolor{red}{\frac{1}{6}} \cdot - \frac{1020}{31}$

$\frac{6}{6} x = - \frac{170}{31}$

$x = - \frac{170}{31}$

The Solution Is: $x = - \frac{170}{31}$ and $y = \frac{56}{31}$ or $\left(- \frac{170}{31} , \frac{56}{31}\right)$