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Question 62171This question is from textbook algebra through applications
: Can you help me solve these equations using the substitution method?
1) x-y=15; y=-4x
2) 5x+3y=0 ;x+y=0
Can you help me solve these equations using the elimination method?
1) a+b=4; -a+2b=-8
2) x+4y=-3; 4x+y=-9
3) 8x+2y=3;x-7y=19
This question is from textbook algebra through applications
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Can you help me solve these equations using the substitution method?
1) x - y = 15; y = -4x
:
Notice they show y = -4x, therefore we can substitute -4x for y in the 1st
equation and solve for x:
x - (-4x) = 15; minus a minus is a plus:
x + 4x = 15
5x = 15
:
x = 15/5, divided both sides by 5
x = 3
:
Find y by substituting 3 for x:
y = -4x
y = -4(3)
y = -12
:
Check our solutions by substituting 3 for x and -12 for y in the 1st equation:
x - y = 15
3 - (-12) = 15
3 + 12 = 15 proves our solutions
:
:
:
2) 5x+3y=0 ;x+y=0
Take the 2nd equation are arrange it in the y = form:
x + y = 0
y = -x, subtracted x from both sides:
:
Substitute -x for y in the 1st equation:
5x + 3y = 0
5x + 3(-x) = 0
5x - 3x = 0
Obviously the only value where this could be true is that x = 0 then y = 0 also
:
:
Can you help me solve these equations using the elimination method?
In this you arrange the equations in a manner where if they are added or subtracted, one of the unknowns is eliminated,
1)
a + b = 4
-a + 2b = -8
------------------adding eliminates a:
0 + 3b = -4
b = -4/3
: Find a, substitute -4/3 for b in the 1st equation:
a + b = 4
a +(-4/3) = 4
a - (4/3) = 4
a = 4 + (4/3); added (4/3) to both sides:
a = (12/3) + (4/3); add using common denominator
a = 16/3
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Check: substitute (-4/3) for b and (16/3) for a in the 2nd equation:
-a + 3b = -8
-(16/3) + 2(-4/3) = -8
-(16/3) - (8/3) = (-24/3) which = -8
:
:
:
2)
x + 4y = -3
4x + y = -9
Mult the 1st equation by 4, leave the 2nd equation as it is:
4x + 16y = -12
4x + y = -9
---------------------Subtracting eliminates x:
0x + 15y = -3
y = -3/15
y = -1/5 or -.2
:
Find x: Substitute =.2 for y in the 1st equation
x + 4y = -3
x + 4(-.2) = -3
x - .8 = -3
x = -3 + .8
x = -2.2
:
Check our solution in the 2nd equation:
4x + y = -9
4(-2.2) + (-.2) = -9
-8.8 - .2 = -9 checks out our solutions:
:
:
3)
8x + 2y = 3
x - 7y = 19
:
Mult the 2nd equation by 8:
8x + 2y = 3
8x -56y = 152
--------------Subtracting eliminates x:
0x + 58y = - 149
y = -149/58; a nasty fraction
:
Find x using the 1st equation
8x + 2y = 3
8x +2(-149/58) = 3
8x - 298/58 = 3
Get rid of the denominator, mult eq by 58:
464x - 298 = 174
464x = 174 + 298
464x = 472
x = 472/464
x = 59/58
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Check using the 2nd equation
x - 7y = 19
(59/58) - 7(-149/58) = 19
(59/58) + 1043/58 = 19 = 19
1102/58 = 19
19 = 19, proves these nasty fractions are indeed the solutions!
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