Systems of equation problems.. Please help!?

Question

Hi,
I'm in intermediate Algebra in college. We are now on chapter 4 which is systems of equations. For the life of me I cannot seem to understand the elimination and substitution process. I'm getting so discouraged with myself over this and am starting to lose the confidence that took me so long to build. Could some please explain these two problems to me in step detail? I really need help. Thanks so much.

1.) x+y=10
y= 4x

2.) 4x-y=9
2x + 3y= -24

Answer 1

1.) by substitution

x+y=10.......1 equation
y=4x............2 equation

2 into 1

x+(4x)=10
5x=10
x=10/5
x=2

when x=2
substitute x=2
y=4(2)
y=8

_________________________________
by elimination

x+y=10.......1 equation
y=4x
4x-y=0..........2 equation

1 plus 2

x+y=10
(+) 4x-y=0
____________
5x=10
x=10/5
x=2

when x=2
y=4(2)
y=8
______________________________

2.) by substitution
4x-y=9
y=4x+9............1 equation
2x+3y=-24.......2 equation

1 into 2

2x+2(4x+9)=-24
2x+8x+18=-24
10x=-24-18
10x=-32
x=-3.2

when x=-3.2
substitute x=-3.2
y=4(-3.2)+9
y=-12.8+9
y=-3.8

_____________________________
by elimination

4x-y=9..........*3
2x+3y=-24

12x-3y=27
(+) 2x+3y=-24
_______________
14x=3
x=3/14

when x=3/14

2x+3y=-24
2(3/14)+3y=-24
3/7+3y=-24
3+21y=-168
21y=-171
y=-171/21

Answer 2

1.)
x + y = 10
y = 4x
Solve first equation for y:
y = 10-x.

Since 4x and 10-x both equal y they are equal:
4x = 10 - x
Add x to both sides:
5x = 10
Divide by 5
x = 2.
Since y = 4x, y = 4(2) = 8

x = 2, y = 8

2.)
4x - y = 9
2x + 3y = -24

Multiply the first equation by 3 to give a -3y that will cancel the 3y from the second equation.

So you now have:
12x - 3y = 27
2x + 3y= -24

Add the second equation to the first:
14x = 3
Divide by 14 and you have
x = 3/14

Plug x in an equation to solve for y.
4x - y = 9
4(3/14) - y = 9
12/14 - y = 9
-y = 114/14
y = - 114/14

x = 3/14, y = -114/14

Answer 3

it's simple
1] y=4x so you substitute y in the first equation by 4x so:
x+4x=10 ......5x=10.....then x=10/5=2
and then y=4x=4*2=8

2]you have solve the two problems together
the first step is to multiply one of the equations with a constant so the coefficiant of x or y is equal in both equations like:
we'll multiply (1) by 3......so 12x-3y=27
and 2x +3y=-24
so if u add the 2 equations the ys will be caceled leaving 14x=3
so x=3/14 and by substituting in any equ. like 2*3/14 +3y=-24
then 3y=-24-3/7
then y=-171/7=-24.43

Answer 4

1) In these equations, we are assuming that x in equation one is the same as x in equation two, and similarly for y. Therefore, if we solve for x or y in equation one and plug it into the other equation, we can solve it.

x + y = 10

Subtracting y from each side give us:

x = 10 - y

Now, we can plug this infor x in the second equation:

y = 4(10 - y)
y = 40 - 4y

Adding 4y to each side of the equation gives:

5y = 40

y = 8

Now, we simply plug this back into either equation:

x + 8 = 10

x = 2


2)The same principle applies here:

4x - y = 9

Let's solve for y in this one. Adding y and subtracting 9 from each side of the equation gives:

4x - 9 = y

Now, we can plug this in for y in the second equation:

2x + 3(4x - 9) = -24
2x + 12x - 27 = -24

Adding 27 to each side of the equation gives us:

14x = 3
x = 3/14

Now, plugging this back into either equation, we can get y:

4(3/14) - y = 9
y = -57/7 or -8-1/7

We can check by plugging x and y back into the other equation

2(3/14) + 3(-57/7) = -24 ?

yes, it works

Answer 5

Question 1
x + 4x = 10
5x = 10
x = 2
y = 8

Question 2
12x - 3y = 27
2x + 3y = - 24-----ADD
14x = 3
x = 3 / 14
6/14 + 3y = - 24
3y = - 24 - 6 / 14
3y = - 24 - 3/7
3y = - 168/7 - 3/7
3y = - 171/7
y = - 57 / 7

Solution: x = 3/14, y = - 57/14

Answer 6

1)
if y = 4x, just substitute y into 4x
like this : x + 4x = 10
5x = 10
x = 2
y = 4x
y = 4*2 = 8

2)
4x-y = 9
2x+3y = -24 --> we can times this equation with 2, so: 4x+6y = -48
and,
4x = 9+y (from first equation)
4x = -48 - 6y (from second equation)
and 4x = 4x
9 + y = -48 -6y
7y = -57
y = -57/7

4x= 9 + y
4x= 6/7
x = 6/28

Answer 7

Nothing to worry.first of all.let us discuss about the substitution method.In your given example there are two variables x and y.
Let us take problem no 1,where the value of y is given as 4x in 2nd equation.So in the first equation of the same problem you put this value in place of y,we are substituting the value of y by x.
so the first equation becomes
x+4x=10
or 5x=10
or,x=2.
So we have already calculated out the value of one variable.now put this value of y in the second equation of the problem and you get the value of the second variable
y.
y=4x=4*2=8
In the second problem.take any one of the equations.For example let us take the first equation which is 4x-y=9
By transpositioning we write it as -y=9-4x
or y=4x-9
Now we have to substitute the value of y in terms of x in the second equation
2x+3(4x-9)= -24
or,2x+12x-27=-24
or,2x+12x= -24+27=3
or,x=3/14
Now we canproceed and find the value of y
Ellimination process is still easier.First of all write the equations one under the another putting them serially
x+y=10.......(eqn1)
y=4x
ory-4x=0
or, -4x+y=0....(eqn 2)
Subtracting eqn(2) fromeqn 1
x+y=10
-4x+y=0
(subtracting) 5x=10
x=10/5=2
now we can get the value of y putting the value of x in any one of the equations