# How do you solve the following linear system: 3x + 5y = -1, 2x − 5y = 16?

Apr 6, 2018

$x = 3 , y = - 2$

#### Explanation:

Starting with the first equation, we need to make either x or y the subject (I chose x)

$3 x + 5 y = - 1$

$x = \frac{- 5 y - 1}{3}$

$2 x - 5 y = 16$

$2 \cdot \frac{- 5 y - 1}{3} - 5 y = 16$

$\frac{- 10 y - 2}{3} - 5 y = 16$

$- 10 y - 2 - 15 y = 48$

$- 25 y = 50$

$y = - 2$

Now we simply substitute $\left(y = - 2\right)$ into the second equation...

$2 x - 5 y = 16$

$2 x - 5 \left(- 2\right) = 16$

$2 x + 10 = 16$

$2 x = 6$

$x = 3$

Finally, we should also verify both answers by substituting both of them into the other equation again.

$3 x + 5 y = - 1$

#3(3) + 5(-2) = -1

$9 - 10 = - 1$

$- 1 = - 1$, so therefore our answers are correct!

Apr 6, 2018

See a solution process below:

#### Explanation:

Step 1) First, solve each equation for $5 y$:

• Equation 1:

$3 x + 5 y = - 1$

$3 x - \textcolor{red}{3 x} + 5 y = - 1 - \textcolor{red}{3 x}$

$0 + 5 y = - 1 - 3 x$

$5 y = - 1 - 3 x$

• Equation 2:

$2 x - 5 y = 16$

$2 x - 5 y + \textcolor{red}{5 y} - \textcolor{b l u e}{16} = 16 - \textcolor{b l u e}{16} + \textcolor{red}{5 y}$

$2 x - 0 - \textcolor{b l u e}{16} = 0 + 5 y$

$2 x - 16 = 5 y$

$5 y = 2 x - 16$

Step 2) Because both equations are equal on the left side we can now equate the right sides of the equation and solve for $x$:

$- 1 - 3 x = 2 x - 16$

$- 1 + \textcolor{b l u e}{16} - 3 x + \textcolor{red}{3 x} = 2 x + \textcolor{red}{3 x} - 16 + \textcolor{b l u e}{16}$

$15 - 0 = \left(2 + \textcolor{red}{3}\right) x - 0$

$15 = 5 x$

$\frac{15}{\textcolor{red}{5}} = \frac{5 x}{\textcolor{red}{5}}$

$3 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}}$

$3 = x$

$x = 3$

Step 3) Substitute $3$ for $x$ in either of the solved equations in Step 1 and solve for $y$:

$5 y = 2 x - 16$ becomes:

$5 y = \left(2 \times 3\right) - 16$

$5 y = 6 - 16$

$5 y = - 10$

$\frac{5 y}{\textcolor{red}{5}} = - \frac{10}{\textcolor{red}{5}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} y}{\cancel{\textcolor{red}{5}}} = - 2$

$y = - 2$

The Solution Is:

$x = 3$ and $y = - 2$

Or

$\left(3 , - 2\right)$

Apr 6, 2018

$x = 3 \mathmr{and} y = - 2$

#### Explanation:

Notice that the terms in $y$ in the two equations are are additive inverses. The sum of additive inverses is $0$
Add the equations together to eliminate the $y$ terms.

$\textcolor{w h i t e}{\times \times \times \times} 3 x \textcolor{b l u e}{+ 5 y} = - 1 \text{ } A$
$\textcolor{w h i t e}{\times \times \times \times} 2 x \textcolor{b l u e}{- 5 y} = 16 \text{ } B$

$A + B : \textcolor{w h i t e}{\times x} 5 x \textcolor{b l u e}{+ 0 y} = 15$
$\textcolor{w h i t e}{\times \times \times \times x} x \textcolor{w h i t e}{\times x} = 3$

Substitute in A:
$\textcolor{w h i t e}{\times \times \times \times} 3 \left(3\right) + 5 y = - 1 \text{ } A$
$\textcolor{w h i t e}{\times \times \times \times \times} 9 + 5 y = - 1$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} 5 y = - 10$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times} y = - 2$

Check in B:
$\textcolor{w h i t e}{\times \times x} 2 \left(3\right) - 5 \left(- 2\right) = 16 \text{ } B$
$\textcolor{w h i t e}{\times \times \times \times \times} 6 + 10 = 16$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} 16 = 16$

The values are correct.