SOLUTION: SO im not sure what these are called but this is what it says... Solve these systems of simultaneous equations using either the substitution method or the addition methed.. check y

Question 193464: SO im not sure what these are called but this is what it says... Solve these systems of simultaneous equations using either the substitution method or the addition methed.. check your work and supply a solution... #1. x+2y=15 and x-2y=-9 #2. x-y=0 and 7x-3y=24 #3. 3x-8y=10 and x-4y=3 #4. 2x+3y=-5 and 3x-y=20.. please help.. the problems are separated by the number sign(#) when it has xxxxx and xxxxx it is with the same problem.. hope that makes sense and hope you can help me.. thanks
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
#1.
x + 2y = 15
x - 2y = -9
-------------note if we add these, we eliminate y, then we can find x easily
2x = 6
x =
x = 3
:
Substitute 3 for x in the 1st equation and find y
3 + 2y = 15
2y = 15 - 3
2y = 12
y =
y = 6
:
Check both solutions in the 2nd equation, substituting for x and y
3 - 2(6) = 12; equality reigns, confirms your solutions
;
:
#2.
x - y = 0
7x -3y =24
From the 1st equation we can see x = y, substitute y for x in the 2nd equation
7y - 3y = 24
4y = 24
y =
y = 6 and x = 6 also
:
Check in 2nd equation
7(6) - 3(6) = 24
42 - 18 = 24
;
:
#3.
3x - 8y = 10
x - 4y = 3
:
Here we can multiply the 2nd by 3 and subtract from the 1st eq, eliminate x
3x - 8y = 10
3x -12y = 9
--------------
0x + 4y = 1
y =
:
Substitute 1/4 for y in the 2nd original equation, find x
x - 4(1/4) = 3
x - 1 = 3
x = 4 + 1
x = 4
:
Check solution in the 1st equation:
3(4) - 8(1/4) = 10
12 - 2 = 10; confirms our solution
:
:
#4.
2x + 3y = -5
3x - y = 20
Use straight substitution here using the 2nd equation
-y = 20 - 3x
Y had to be positive, multiply both sides by -1
y = 3x - 20
:
Substitute (3x-20) for y in the 1st equation
2x + 3(3x+20) = -5
2x + 9x - 60
11x = -5 + 60
11x = 55
x =
x = 5
:
we said y = 3x - 20
y = 3(5) - 20
y = 15 - 20
y = -5
:
Why don't you check these solutions in both equations

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