How do you solve the system $5x= -25+5y$ and $10y=42+2x$ ?

Question

3561 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-13T22:30:15

Divide the first equation by $5$ to get an expression for $x$ to substitute in the second equation. Solve to find $y=4$ and hence $x=-1$

Divide both sides of the first equation by $5$ to get:

$x = -5+y = y-5$

Substitute $y-5$ for $x$ in the second equation:

$10y = 42 + 2x = 42 + 2(y - 5)$

$=42+2y - 10 = 32+2y$

Subtract $2y$ from both ends to get:

$8y = 32$

Divide both sides by $8$ to get:

$y = 4$

Then

$x = y-5 = 4-5 = -1$

How do you solve the system 5x= -25+5y5x= -25+5y and 10y=42+2x10y=42+2x?