SOLUTION: 1.
3x=-5-x
2x+y=-5
2.
3x-5y=7
2x-y=-7
3.
x-3y=1
3x-5y=-5
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Question 168738: 1.
3x=-5-x
2x+y=-5
2.
3x-5y=7
2x-y=-7
3.
x-3y=1
3x-5y=-5
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1
Your first equation has two "x" variables. Is there a "y" term in the first equation?
# 2
Start with the given system of equations:
Multiply the both sides of the second equation by -5.
Distribute and multiply.
So we have the new system of equations:
Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:
Group like terms.
Combine like terms. Notice how the y terms cancel out.
Simplify.
Divide both sides by to isolate .
Reduce.
------------------------------------------------------------------
Now go back to the first equation.
Plug in .
Multiply.
Add to both sides.
Combine like terms on the right side.
Divide both sides by to isolate .
Reduce.
So our answer is and .
Which form the ordered pair )
.
This means that the system is consistent and independent.
Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer.
Graph of (red) and (green)
# 3
Start with the given system of equations:
Multiply the both sides of the first equation by -3.
Distribute and multiply.
So we have the new system of equations:
Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:
Group like terms.
Combine like terms. Notice how the x terms cancel out.
Simplify.
Divide both sides by to isolate .
Reduce.
------------------------------------------------------------------
Now go back to the first equation.
Plug in .
Multiply.
Add to both sides.
Combine like terms on the right side.
Divide both sides by to isolate .
Reduce.
So our answer is and .
Which form the ordered pair )
.
This means that the system is consistent and independent.
Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer.
Graph of (red) and (green)
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