Algebra please help?

Question

1.Solve the system by substitution.
x + y = 12
y = 2x
2.Solve the system by addition.
x + y = 4
x – y = 4
3.Rewrite the equation y = –4x + 3 as a function of x.

Answer 1

1. Since y = 2x, replace the y in eq. 1 with 2x to get x + 2x = 12

3x = 12

solve for x then plug it in to get y.


2. Add the two equations together.
That comes out 2x + 0y = 8, or just 2x = 8.

Solve for x, and plug in to either equation to get y.


3. Subtract 3 from both sides, then divide both sides by -4

Answer 2

Here is how you solve the first one:

1. y=2x, so you plug y into the equation x+y=12
2. you will get x+2x=12 (you substitute the y with the 2x)
3. This will be 3X=12
4. X=4
5. Solve for Y, which is 8

The second one

1. At first glance, the "Y" will cancel out because one is a negative and the other is a positive.
2. You add the two X's together and the two "4s" together
3. You get 2X=8
4. X=4
5. Plug in the X into either equation and solve for Y
6. Y should equal 0

The third one

1. Move the 3 to the other side (Y-3=-4X)
2. Divide both sides by -4
3. The answer is X= -(Y-3)/4

Answer 3

From equn.no 1 x+y=12 ,we get x=12-y. Substituting the value of x in equn.no.2 which is y=2x,we get y=2(12-y) =>y=24-2y =>y+2y=24 or 3y=24 =>y=24/3=8 .Putting the value of y in equn.no1,we get x+8=12 =>x=12-8=4 (2) Adding the two equations,we get 2x=8 => x=8/2=4 Putting the value of x in equn no 1.we get 4+y=4 =>y=4-4=0 (3) y=-4x+3 => -4x+3 =y =>-4x=y-3 => 4x=3-y => x=(3-y)/4

Answer 4

1.
subst y for 2x
x + 2x = 12
3x = 12
x = 4

2.
add the 2 equations
x + y= 4
+
x - y = 4
2x = 8
x = 4
subst x=4 in one of the equations
x - y= 4
4 - y = 4
-y = 0
y = 0

3.
make the equation equal x
4x = 3 - y
x = 3 - y divided by 4


Glad to help

Answer 5

x+2x=12
3x=12
x=4
y=8


2x=8
x=4

y=0

(y-3)/-4=x

j