Solve the system?

Question

1. Solve the system by substitution.
2x – 2y = –2
y = 5x – 19


2. Solve the system by addition.
5x – 3y = 13
4x – 3y = 11


3. Solve the system by substitution.
x + y = 12
y = 2x


Algebra Help.

Thank you.

Answer 1

1. x = 5, y = 6
2. x = 2, y = -1
3. x = 4, y = 8

1. Substitute for y in the first equation: 2x - 2*(5x - 19) = -2 and expand it all to get: 2x - 10x + 38 = -2, then combine the x's and subtract 38 from each side: -8x + 38 = -2 - 38 yielding: -8x = -40 and finally divide by -8 to get x = -40/-8 = 5. The substitute into either (the 2nd is faster) equation with x = 5 to get: y = 5*5 -19 = 25 - 19 = 6.

2. To solve using addition, let's note the -3y in each equation and figure what to do to one of them (oh, the 2nd one) so they will go away when added. Multiplying by -1 is what to do: -1*(4x - 3y) = -1*(11) getting: -4x + 3y = -11. Now, add the two equations: 5x - 3y + (-4x + 3y) = 13 + (-11) and combine terms like so: 5x - 4x - 3y + 3y = 13 -11 and reduce the confusion to: x = 2. Now substitute back into either one, the 1st perhaps: 5*2 - 3y = 13 and subtract the 5*2 (=10) from each side for: -3y = 3 and finally divide by -3 on each side to get: y = -1.

3. Ah, substitution again! Put the y = 2x to work in the 1st equation: x + (2x) = 12 and combine: 3x = 12 and finally divide each side by 3: x = 4. Now substitute back into either equation (we'll take the 1st): 4 + y = 12 and subtract 4 from each side to get: y = 8.

Answer 2

Substitution means that you find the 'value' of one variable in one equation, then use that value for the variable in the second equation.

from 2x - 2y = -2
you get x = y - 1

In the second equation, use (y-1) instead of x
y = 5(y-1) - 19
y = 5y - 5 - 19
24 = 4y (and so on)

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2 Addition: multiply one (or both) equation(s) throughout by a constant so that one variable's factor will cancel out.
Here, we can multiply the top one by -1 throughout, then add it to the second equation:
-5x + 3y = -13
+4x - 3y = +11

-x + 0 y = -2
-x = -2
(x = 2)

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Answer 3

1. Solve the system by substitution.
2x – 2y = –2
y = 5x – 19

2x - 2*(y) = -2
2x - 2* (5x-19) = -2
2x - 10x +38 = -2
-8x = 40
x = -5

y = 5*x - 19
y = 5*(-5) - 19
y = -25 - 19
y = -44


2. Solve the system by addition.
5x – 3y = 13
4x – 3y = 11

5x-3y = 13
-4x+3y = -11 (+)
--------------
x = 2


20x - 12y = 52
-20x + 15y = -55 (+)
------------------
3y = -3
y = -1

3. Solve the system by substitution.
x + y = 12
y = 2x

x + y = 12
x + (2x) = 12
3x = 12
x = 4

y = 2x
y = 2*4
y = 8

Answer 4

1) sub y =5x-19 into the 1st eqn,
2x-2(5x-19)=-2
2x-10x+38=-2
-8x=-2-38
-8x=-40
x=5, y=5(5)-19=6

2) 5x – 3y = 13--->eqn1
4x – 3y = 11--->eqn2

Use eqn1 minus eqn2,
5x-3y-(4x-3y)=13-11
x+0y=2
sub x=2 into eqn1,
5(2)-3y=13
10-3y=13
3y=-3
y=-1

3) sub y = 2x into eqn 1,
x+2x=12
3x=12
x=4
y=2*4=8

Answer 5

1)
x=5
y=6
2)
x=2
y=-1
3)
x=4
y=8

Answer 6

2x-2(5x-19)=-2
2x-10x+38=-2
-8x=-40
x=5

2(5)-2y=-12
-2y=-12
y=6

--------------
5x-3y=13
-(4x-3y=11)

x=2

5(2)-3y=13
y=-1
------------------
x+y=12
y=2x

x+2x=12
3x=12
x=4

4+y=12
y=8

voilá!!!