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

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Question 168732: solve each system by any method
1 x-3y=1
3x-5y=-5
2 2x-3y=5
4x-6y=3
: solve each system by any method
1 x-3y=1
3x-5y=-5
2 2x-3y=5
4x-6y=3

Answer by jim_thompson5910(9911) About Me  (Show Source):
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# 1

Substitution:

x-3y=1 Start with the first equation


x=3y+1 Add 3y to both sides.


3x-5y=-5 Move onto the second equation


3(3y+1)-5y=-5 Plug in x=3y+1


9y+3-5y=-5 Distribute.


4y+3=-5 Combine like terms on the left side.


4y=-5-3 Subtract 3 from both sides.


4y=-8 Combine like terms on the right side.


y=(-8)/(4) Divide both sides by 4 to isolate y.


y=-2 Reduce.


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x=3y+1 Go back to the first isolated equation


x=3(-2)+1 Plug in y=-2


x=-6+1 Multiply


x=-5 Add



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So the solutions are x=-5 and y=-2 which form the ordered pair (-5, -2)







# 2

Elimination:




Start with the given system of equations:
system(2x-3y=5,4x-6y=3)


-2(2x-3y)=-2(5) Multiply the both sides of the first equation by -2.


-4x+6y=-10 Distribute and multiply.


So we have the new system of equations:
system(-4x+6y=-10,4x-6y=3)


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


(-4x+6y)+(4x-6y)=(-10)+(3)


(-4x+4x)+(6y+-6y)=-10+3 Group like terms.


0x+0y=-7 Combine like terms. Notice how the x terms cancel out.


0=-7Simplify.


Since 0=-7 is never true, this means that there are no solutions. So the system is inconsistent.