# How do you solve the simultaneous equations 3x+7y=26 and 4x+5y=13?

Jul 18, 2015

I found:
$x = - 3$
$y = 5$

#### Explanation:

I would multiply the first equation by $- 4$ and the second by $3$ and add together the two equatins (in columns):
{-12x-28y=-104
{12x+15y=39
$0 - 13 y = - 65$
So: $y = \frac{65}{13} = 5$
Substitute back this value into the first equation:
$3 x + 7 \cdot 5 = 26$
$3 x = 26 - 35 = - 9$
$x = - \frac{9}{3} = - 3$

Jul 18, 2015

$\textcolor{red}{x = - 3 , y = 5}$ or $\textcolor{red}{\text{(-3, 5)}}$

#### Explanation:

One way is to use the method of elimination.

Step 1. Enter the equations.

Equation (1) $3 x + 7 y = 26$
Equation (2) $4 x + 5 y = 13$

Step 2. Prepare the equations.

Multiply every term in each equation by a number so that equal terms can be eliminated.

Multiply Equation (1) by $4$ and Equation (2) by $3$.

Equation (3) $12 x + 28 y = 104$
Equation (4) $12 x + 15 y = 39$

Step 3. Subtract Equation (4) from Equation (3).

(3) $12 x + 28 y = 104$
(4) $12 x + 15 y = 39$
(5) $\overline{\textcolor{w h i t e}{000000} 13 y = 65}$

$y = \frac{65}{13}$

Equation (6) $y = 5$

Step 4. Substitute Equation (6) in Equation (2).

$4 x + 5 y = 13$
4x+5×5 =13
$4 x + 25 = 13$
$4 x = 13 - 25$
$4 x = - 12$
$x = - \frac{12}{4}$

$x = - 3$

Solution: $x = - 3 , y = 5$ or $\left(- 3 , 5\right)$

Check: Substitute the values of $x$ and $y$ in Equation (1).

3x+7y = 3(-3) + 7×5 = -9+35= 26