How do you solve #3x+3y=15# and #2x+3y=-5#?

Question

Kennedy Adams · Accepted Answer

The solution for the system of equations is
#color(blue)(x=20,y=-15# #3x+color(blue)(3y)=15 #.....equation #(1)# #2x+color(blue)(3y)=−5#....equation #(2)# Solving by elimination Subtracting equation #2# form #1# results in elimination of #color(blue)(3y#: #3x+cancelcolor(blue)(3y)=15 # #-2x-cancelcolor(blue)(3y)=+5# #x=color(blue)(20# Substituting #x# in equation #1# to find #y# #3y=15-3x # #3y=15-3*(20) # #3y=15-60 # #y=-45/3 # #color(blue)(y=-15#

Julia Adams · Answer

undefined To solve the system of equations $3x + 3y = 15$ and $2x + 3y = -5$, you can use the method of substitution or elimination. Let's use the elimination method:

1. Multiply the second equation by 3 to make the coefficients of $y$ the same in both equations:
$$6x + 9y = -15$$

2. Subtract the second equation from the first equation:
$$3x - 6x + 3y - 9y = 15 - (-15)$$
$$ -3x = 30$$

3. Divide both sides by -3:
$$x = -10$$

4. Substitute $x = -10$ into either of the original equations. Let's use the first equation:
$$3(-10) + 3y = 15$$

5. Solve for $y$:
$$-30 + 3y = 15$$
$$3y = 45$$
$$y = 15$$

So the solution to the system of equations is $x = -10$ and $y = 15$.