How do you solve the following system of equations algebraically: 3x + 2y = 4, 4x + 3y = 7?

Nov 17, 2016

Substitute one equation into the other to give $x = - 2$, and $y = 5$

Explanation:

You have two equations in two unknowns, $x$ and $y$, so this system is potentially solvable exactly.

$3 x + 2 y = 4$ $\left(i\right)$

$4 x + 3 y = 7$ $\left(i i\right)$

From $\left(i\right)$, $x = \frac{4 - 2 y}{3}$, so substitute this value into $\left(i i\right)$:

$4 \left(\frac{4 - 2 y}{3}\right) + 3 y = 7$

$\frac{16}{3} - \frac{8 y}{3} + 3 y = 7$; multiply thru by $3$:

$16 - 8 y + 9 y = 21$

$y = 21 - 16 = 5$

And substitute this value back into $\left(i\right)$:

$3 x + 2 \left(5\right) = 4$, $x = - 2$

If we recheck the original equations, these values are correct.