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

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Answer 1 · 2015-08-16T02:01:31

$The solution is (2,1).$ $color(red)("The solution is (2,1).")$

Step 1. Enter the equations.

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

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

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

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

Step 3. Subtract Equation 4 from Equation 3.

[5] $y = 1$ $y=1$

Step 4. Substitute Equation 5 in Equation 1.

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

$x = 2$ $x=2$

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

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

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

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

They check!

Our solution is correct.