Question

Solve a) with Substitution and b) with Elimination (check attached picture)

Answer 1

a) Solve the first equation for $x$.

$x + 3y = 7 \impiles x = 7 - 3y$

Substitute this into the second equation and solve for $y$.

$2x + 4y = 12 \implies 2(7 - 3y) + 4y = 12 \\\\ \implies 14 - 6y + 4y = 12 \\\\ \implies -2y = -2 \\\\ \implies \boxed{y=1}$

Solve for $x$.

$x = 7 - 3y \implies x = 7-3\times1 \\\\ \implies \boxed{x = 4}$

b) Eliminate $y$ by combining the two equations in appropriate parts, namely

$2 (2x - y) + (3x + 2y) = 2\times3 + (-3)$

and solve for $x$.

$2 (2x - y) + (3x + 2y) = 2\times3 + (-3) \implies 4x - 2y + 3x + 2y = 6 - 3 \\\\ \implies 7x = 3 \\\\ \implies \boxed{x = \dfrac37}$

Solve for $y$.

$2x - y = 3 \implies 2\times\dfrac37 - y = 3 \\\\ \implies \dfrac67 - y = 3 \\\\ \implies -y = \dfrac{15}7 \\\\ \implies \boxed{y = -\dfrac{15}7}$