Question

1. What is the solution to the system of equations?

Answer 1

Answer:

$\large\boxed{x=-8\ and\ y=9\to(-8,\ 9)}$

Step-by-step explanation:

$a)\ \text{Elimination method:}\\\\\left\{\begin{array}{ccc}3x+4y=12\\x+2y=10&\text{multiply both sides by (-2)}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}3x+4y=12\\-2x-4y=-20\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad \boxed{x=-8}\\\\\text{Put the value of x to the second equation:}\\-8+2y=10\qquad\text{add 8 to both sides}\\2y=18\qquad\text{divide both sides by 2}\\\boxed{y=9}$

$b)\ \text{Substitution method:}\\\\\left\{\begin{array}{ccc}3x+4y=12\\x+2y=10&\text{subtract 2y from both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}3x+4y=12&(1)\\x=10-2y&(2)\end{array}\right\qquad\text{subtract (2) to (1)}\\\\3(10-2y)+4y=12\qquad\text{use distributive property}\\(3)(10)+(3)(-2y)+4y=12\\30-6y+4y=12\qquad\text{subtract 30 from both sides}\\-2y=-18\qquad\text{divide both sides by (-2)}\\\boxed{y=9}\\\\\text{Put the value of y to (2):}\\x=10-2(9)\\x=10-18\\\boxed{x=-8}$