Solve Systems of Equations by Elimination: example 5.31

Solve the system by elimination. ${\begin{matrix} −6 x + 15 y = 10 \\ 2 x - 5 y = −5 \end{matrix}$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the system by elimination. ${\begin{matrix} −3 x + 2 y = 8 \\ 9 x - 6 y = 13 \end{matrix}$

How did it go?

The equations are in standard form.	${\begin{matrix} - 6 x + 15 y & = & 10 \\ 2 x - 5 y & = & −5 \end{matrix}$
Multiply the second equation by 3 to eliminate a variable.	${\begin{matrix} - 6 x + 15 y & = & 10 \\ 3 (2 x - 5 y) & = & 3 (−5) \end{matrix}$
Simplify and add.	$\begin{matrix} \underset{__________________}{{\begin{matrix} - 6 x + 15 y & = & 10 \\ 6 x - 15 y & = & −15 \end{matrix}} \\ 0 \neq −5 \end{matrix}$

The equations are in standard form.	${\begin{matrix} - 6 x + 15 y & = & 10 \\ 2 x - 5 y & = & −5 \end{matrix}$
Multiply the second equation by 3 to eliminate a variable.	${\begin{matrix} - 6 x + 15 y & = & 10 \\ 3 (2 x - 5 y) & = & 3 (−5) \end{matrix}$
Simplify and add.	$\begin{matrix} \underset{__________________}{{\begin{matrix} - 6 x + 15 y & = & 10 \\ 6 x - 15 y & = & −15 \end{matrix}} \\ 0 \neq −5 \end{matrix}$