substitution algebra help plz?

Question

Thank you!

I just dont get it my teacher is horrible!

these are the qustiones i need to do and there so hard. He didnt even explain hot to do them just gave us a wk sheet and boook work =(

can someone plz help me and describe how to solve these?

Solve eacg equation for x, then solve each equation for y.

3x-5y=12


use substituaion to solve each sytem of equationes. If the system doe snot have exaclty one solution, state wheter it has no solution or infinite solutions.

1.2x+7y=3
x=1-4y


2. 2x+3y=1
-3x+y=15


3. x=1/2y+3
2x-y=6

4. y=3/5x
3x-5y=15



thx so much!

Answer 1

3x= 12 + 5y
x = (12 + 5y) / 3
5y = 3x - 12
y = (3x - 12) / 5

Question 1
2x + 7y = 3
- 2x - 8y = - 2---ADD
- y = 1
y = - 1
x = 1 - 4y
x = 1 + 4
x = 5
Solution is x = 5 , y = - 1

Question 2
6x + 9y = 3
- 6x + 2y = 30---ADD
11 y = 33
y = 3
2x + 9 = 1
2x = - 8
x = - 4
Solution is x = - 4 , y = 3

Question 3
- 2x + y = - 6
2x - y = 6
Error in question

Question 4
3x - 3x = 0
Error in question

Answer 2

What you're trying to do is isolate x or y on one side of one equation and fit that into the other equation. It doesn't matter whether you solve x or y first.

Taking problem 1, they make it easy for you. x is already solved, so lets plug it into the first equation.
x = (1-4y)
2(1-4y) + 7y =3
simplify
2-8y+7y = 3
2-y = 3
-y = 1
y = -1
plug this back into either of the equations
x = 1-4(-1)
x = 1+4
x = 5

Problem 2, we need to isolate one of the variables. Lets use the second one for simplicity sake:
-3x+y = 15
add 3x to each side
-3x+y+3x = 15 + 3x
y = 15+3x
plug this y into the first equation
2x + 3(15+3x) = 1
2x + 45 +9x =1
11x +45 = 1
subtract 45 from both sides
11x = -44
x = -4
Plug it back into any equation
-3(-4) + y = 15
12 + y = 15
y = 3

Problem 3
2(1/2y + 3) -y = 6
1/y + 6 - y = 6
1/y -y = 0
multiply both sides by y
1-y^2 = 0
-y^2 = -1
y^2 = 1
y = 1
2x-1=6
2x = 7
x = 7/2 or 3.5

4)
3x - 5(3/5x) = 15
3x - 3/x= 15
multiply by x on both sides
3x^2 - 3 = 15 x
3x^2 - 15x -3 = 0
divide by 3 on both sides
x^2 - 5x - 1 = 0
The answer is difficult, x = (5+- squareroot(29)) / 2
So, I think the question is misworded.

Answer 3

3x-5y=12

Solve for x

3x = 5y + 12 {add 5y to both sides}
x = 5y/3 + 4 {divide by 3}

Solve for y

-5y = -3x + 12 {subtract 3x}
y = 3x/5 - 12/5 {divide by -5}


1a) 2x+7y=3
1b) x=1-4y

The substitution method when solving a system of two equations in two unknowns: solve one of the equations for one of the variables and then replace that variable in the other equation with the solution so you have one equation in one unknown

Here equation 1b is already solved for x, so substitute into equation 1a

1a) 2x+7y=3
2(1 - 4y) + 7y = 3
2 - 8y + 7y = 3
-y = 1
y = -1

Plug this result into equation 1b

1b) x=1-4y
x = 1 - 4(-1)
x = 1 + 4
x = 5

Check using equation 1a

1a) 2x+7y=3
2(5) + 7(-1) = 3
10 - 7 = 3
3 = 3

Solution is (5, -1)


2a) 2x+3y=1
2b) -3x+y=15

Equation 2b can readily be solved for y

2b) -3x + y = 15
y = 3x + 15

Substitute this for y in equation 2a and continue as above


3a) x=1/2y+3
3b) 2x-y=6

1/2y is ambiguous

1/2 of y is best written as (1/2)y or y/2

Problems 3 and 4 are no more difficult than the others, except you need to be careful with the fractions

Answer 4

These are pretty simple actually... take no.1 for example. Step one is to take one of the two equations (whichever is easier) and solve for either x or y. In no.1, this is already done. Then plug that in to the other equation. Like so...

2x+7y=3
x=1-4y
------------
2(1-4y)+7y=3

Then solve for y.

2-8y+7y=3 so y=-1

then plug y=(-1) back into either of the two original equations.

x=1-4(-1) so x=5

you check yourself by plugging both x and y into the other equation and make sure its equal.

Answer 5

for q1
put x=1-4y in first equation
2(1-4y)+7y=3
2-8y+7y=3
-y=3-2
-y=1
y=-1
now put y in equation2
x=1-4(-1)
=1+4
x=5