# How do you solve the system using the elimination method for 2x+3y=7 and 3x+4y=10?

Aug 16, 2015

$\textcolor{red}{\text{The solution is (2,1).}}$

#### Explanation:

Step 1. Enter the equations.

[1] $2 x + 3 y = 7$
[2] $3 x + 4 y = 10$

Step 2. Multiply each equation by a number to get the lowest common multiple for the coefficients of one variable.

Multiply Equation 1 by $3$ and Equation 2 by $2$.

[3] $6 x + 9 y = 21$
[4] $6 x + 8 y = 20$

Step 3. Subtract Equation 4 from Equation 3.

[5] $y = 1$

Step 4. Substitute Equation 5 in Equation 1.

$2 x + 3 y = 7$
2x+3×1 =7
$2 x + 3 = 7$
$2 x = 4$

$x = 2$

Solution: The solution that satisfies both equations is $\left(2 , 1\right)$.

Check: Substitute the values of $x$ and $y$ in each Equation.

$2 x + 3 y = 7$
2×2+3×1=7
$4 + 3 = 7$
$7 = 7$

$3 x + 4 y = 10$
3×2+4×1=10
$6 + 4 = 10$
$10 = 10$

They check!

Our solution is correct.