Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 152235: HOW WOULD I SOLVE THIS algebracially ?

3/4x-2/5y=2
1/2x-3/5y=-2
: HOW WOULD I SOLVE THIS algebracially ?

3/4x-2/5y=2
1/2x-3/5y=-2

Answer by jim_thompson5910(9162) About Me  (Show Source):
You can put this solution on YOUR website!

(3/4)x-(2/5)y=2 Start with the first equation.


20((3/cross(4))x-(2/cross(5))y)=20(2) Multiply both sides by the LCD 20 to clear any fractions.


15x-8y=40 Distribute and multiply.


(1/2)x-(3/5)y=-2 Move onto the second equation.


10((1/cross(2))x-(3/cross(5))y)=10(-2) Multiply both sides by the LCD 10 to clear any fractions.


5x-6y=-20 Distribute and multiply.


So we have the system of equations:


system(15x-8y=40,5x-6y=-20)



-3(5x-6y)=-3(-20) Multiply the both sides of the second equation by -3.


-15x+18y=60 Distribute and multiply.


So we have the new system of equations:
system(15x-8y=40,-15x+18y=60)


Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:


(15x-8y)+(-15x+18y)=(40)+(60)


(15x-15x)+(-8y+18y)=40+60 Group like terms.


0x+10y=100 Combine like terms. Notice how the x terms cancel out.


10y=100 Simplify.


y=(100)/(10) Divide both sides by 10 to isolate y.


y=10 Reduce.


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15x-8y=40 Now go back to the first equation.


15x-8(10)=40 Plug in y=10.


15x-80=40 Multiply.


15x=40+80 Add 80 to both sides.


15x=120 Combine like terms on the right side.


x=(120)/(15) Divide both sides by 15 to isolate x.


x=8 Reduce.


So our answer is x=8 and y=10.


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.


drawing(500,500,-2,18,-2,20,<BR>
grid(1),<BR>
graph(500,500,-2,18,-2,20,(40-15x)/(-8),(-20-5x)/(-6)),<BR>
circle(8,10,0.05),<BR>
circle(8,10,0.08),<BR>
circle(8,10,0.10)<BR>
) Graph of (3/4)x-(2/5)y=2 (red) and (1/2)x-(3/5)y=-2 (green)