SOLUTION: I need to solve the systems of equations by elimination and I do not understand how to solve by substitution. Can you help? y+5x=-3 3y-2x=8 Thank you so much!

Algebra ->  Equations -> SOLUTION: I need to solve the systems of equations by elimination and I do not understand how to solve by substitution. Can you help? y+5x=-3 3y-2x=8 Thank you so much!      Log On


   

Question 901185: I need to solve the systems of equations by elimination and I do not understand how to solve by substitution. Can you help?
y+5x=-3
3y-2x=8
Thank you so much!
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
y%2B5x=-3
3y-2x=8
or
5x%2By=-3
-2x%2B3y=8

Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

5%2Ax%2B1%2Ay=-3
-2%2Ax%2B3%2Ay=8

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 5 and -2 to some equal number, we could try to get them to the LCM.

Since the LCM of 5 and -2 is -10, we need to multiply both sides of the top equation by -2 and multiply both sides of the bottom equation by -5 like this:

-2%2A%285%2Ax%2B1%2Ay%29=%28-3%29%2A-2 Multiply the top equation (both sides) by -2
-5%2A%28-2%2Ax%2B3%2Ay%29=%288%29%2A-5 Multiply the bottom equation (both sides) by -5


So after multiplying we get this:
-10%2Ax-2%2Ay=6
10%2Ax-15%2Ay=-40

Notice how -10 and 10 add to zero (ie -10%2B10=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%28-10%2Ax%2B10%2Ax%29-2%2Ay-15%2Ay%29=6-40

%28-10%2B10%29%2Ax-2-15%29y=6-40

cross%28-10%2B10%29%2Ax%2B%28-2-15%29%2Ay=6-40 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

-17%2Ay=-34

y=-34%2F-17 Divide both sides by -17 to solve for y



y=2 Reduce


Now plug this answer into the top equation 5%2Ax%2B1%2Ay=-3 to solve for x

5%2Ax%2B1%282%29=-3 Plug in y=2


5%2Ax%2B2=-3 Multiply



5%2Ax=-3-2 Subtract 2 from both sides

5%2Ax=-5 Combine the terms on the right side

cross%28%281%2F5%29%285%29%29%2Ax=%28-5%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5 on the left side.


x=-1 Multiply the terms on the right side


So our answer is

x=-1, y=2

which also looks like

(-1, 2)

Notice if we graph the equations (if you need help with graphing, check out this solver)

5%2Ax%2B1%2Ay=-3
-2%2Ax%2B3%2Ay=8

we get



graph of 5%2Ax%2B1%2Ay=-3 (red) -2%2Ax%2B3%2Ay=8 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (-1,2). This verifies our answer.




Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++5%5Cx+%2B+1%5Cy+=+-3%2C%0D%0A++++-2%5Cx+%2B+3%5Cy+=+8+%29%0D%0A++We'll use substitution. After moving 1*y to the right, we get:
5%2Ax+=+-3+-+1%2Ay, or x+=+-3%2F5+-+1%2Ay%2F5. Substitute that
into another equation:
-2%2A%28-3%2F5+-+1%2Ay%2F5%29+%2B+3%5Cy+=+8 and simplify: So, we know that y=2. Since x+=+-3%2F5+-+1%2Ay%2F5, x=-1.

Answer: system%28+x=-1%2C+y=2+%29.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I need to solve the systems of equations by elimination::
y+5x=-3
3y-2x=8
------------
Modify for elimination::
3y + 15x = -9
3y - 2x = 8
----------
Subtract and solve for "x":
17x = -17
x = -1
------
Solve for "y" using y + 5x = -3
y + 5(-1) = -3
y - 5 = -3
y = 2
------
Ans: (-1,2)
===================
Cheers,
Stan H.
-------------------