How do you solve the system $3 y = - \frac{1}{2} x + 2$ $3y=-1/2x+2$ and $y = - x + 9$ $y=-x+9$ ?

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Answer 1 · 2015-06-14T23:06:01

Substitute the expression for $y$ $y$ from the second equation into the first and solve for $x$ $x$ to get $x = 10$ $x = 10$ and $y = - 1$ $y = -1$

Substitute the expression for $y$ $y$ from the second equation into the first...

$- \frac{1}{2} x + 2 = 3 y = 3 (- x + 9) = - 3 x + 27$ $-1/2x+2 = 3y = 3(-x+9) = -3x+27$

Add $3 x$ $3x$ to both sides to get:

$\frac{5}{2} x + 2 = 27$ $5/2x+2 = 27$

Subtract $2$ $2$ from both sides to get:

$\frac{5}{2} x = 25$ $5/2x = 25$

Multiply both sides by $\frac{2}{5}$ $2/5$ to get:

$x = 10$ $x = 10$

Then $y = - x + 9 = - 10 + 9 = - 1$ $y = -x+9 = -10+9 = -1$

How do you solve the system 3y=-1/2x+23y=−12x+23y=-1/2x+2 and y=-x+9y=−x+9y=-x+9?