How do you solve #3x-y=9# and #2x+y=1#?

Question

How do you solve #3x-y=9# and  #2x+y=1#?

Julia Adams · Accepted Answer

The solution for the system of equations is
#color(blue)(x=2,y=-3# #3x-color(blue)(y)=9#......equation #(1)# #2x +color(blue)(y)=1#......equation #(2)# Solving by elimination Adding equations #1# and #2# #3x-cancelcolor(blue)(y)=9# #2x +cancelcolor(blue)(y)=1# #5x =10# #x=10/5# #color(blue)(x=2# Finding #y# from equation #1# #3x-y=9# #3 xx 2 -y=9# #6 -y=9# #6 -9=y# #color(blue)(y=-3#

Hazel Anglin · Answer

undefined To solve the system of equations $3x - y = 9$ and $2x + y = 1$:

1. Solve one of the equations for one variable.
2. Substitute the expression for that variable into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found back into one of the original equations to solve for the other variable.
5. Check the solution by verifying that it satisfies both equations.

Solving for y in the first equation:
$$3x - y = 9$$
$$y = 3x - 9$$

Substituting $3x - 9$ for $y$ in the second equation:
$$2x + (3x - 9) = 1$$

Solving for $x$:
$$2x + 3x - 9 = 1$$
$$5x - 9 = 1$$
$$5x = 10$$
$$x = 2$$

Substituting $x = 2$ into $y = 3x - 9$:
$$y = 3(2) - 9$$
$$y = 6 - 9$$
$$y = -3$$

So, the solution to the system of equations is $x = 2$ and $y = -3$.