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Question 102566: Use the elimination method to find all solutions of the system;
A. 5x+2y=-19
7x+3y=-27
x=___ y=____
b. x+3y=5
6y+z=12
x-2x=10
c. 2/x+3y=16
-1/x+2/y=6
d. y=64-x^2
y=x^2-64
The two solutions of the system are:
one with x<0 x=___ y=____
one with x>0 x=___ y=____
I have tried working the elimination rules but am confused with getting rid of x in the second equation to solve for y?
Answer by rmromero(383)   (Show Source):
You can put this solution on YOUR website!
Use the elimination method to find all solutions of the system;
A. 5x+2y=-19
7x+3y=-27
x=___ y=____
Let us eliminate x to solve for y. the equations are:
5x + 2y = -19 eqn 1
7x + 3y = -27 eqn 2
To eliminate x, multiply -7 to eqn 1 and 5 to eqn 2.
-7 (5x + 2y = -19)
5 (7x + 3y = - 27)
-35x - 14y = 133 eqn 1
35x + 15y = -135 eqn 2
Add eqn 1 and eqn 2, answer is:
-35x - 14y = 133 eqn 1
35x + 15y = -135 eqn 2
__________________________
0 - y = -2
Therefore y = -2. Substitute y = -2 to eqn 1 or eqn 2 to find x.
5x + 2y = -19, y = -2
5x + 2(-2) = -19
5x - 4 = - 19 Add 4 both sides.
5x - 4 + 4 = -19 + 4
5x = -15 Divide both sides by 5 so x will be left.
x = -3
Therefore x = -3 and y = -2
You try b - d.
b. x+3y=5
6y+z=12
x-2x=10
Find the simpliest eqn first. look at x - 2x = 10. with this you can
solve for x.
x - 2x = 10
-x = 10 divide -1 both sides to make x positive.
x = -10
Since you have the value of x, you can substitute it to the eqn x + 3y = 5
x + 3y = 5, where x = -10
-10 + 3y = 5 Add 10 both sides
-10 + 10 + 3y = 5 + 10
3y = 15 Divide both sides by 3
y = 5
Now you can solve for z using y = 5 and the eqn 6y+z=12.
6y + z = 12 , where y = 5
6(5) + z = 12
30 + z = 12 Subtract both sides by 30
30 - 30 + z = 12 - 30
z = -18
Therefore x = -10, y = 5 and z = -18
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