Hi, I am taking Algebra 2 and I need help with some Substitution method problems:

x+12y=68
x=8y-12

x+3y=7
2x-4y=24

3a+b=3
2a-5b=-15

-6=3x-6y
4x=4+5y

I'll do one for you. They all work the same way. You need the practice doing them yourself.

Substitute 8y - 12 for x in the first equation.
8y - 12 + 12y = 68
20 y = 80
y = 4
x = 8y - 12 = 32 - 12 = 20

Sure, I can help you with substitution method problems in Algebra 2. Let's go through each problem step by step.

Problem 1:
x + 12y = 68
x = 8y - 12

To solve this system of equations using the substitution method, we'll solve one equation for a variable and substitute it into the other equation.

From the second equation, x = 8y - 12. We can substitute this expression for x in the first equation:

(8y - 12) + 12y = 68

Combine like terms:
20y - 12 = 68

Add 12 to both sides:
20y = 80

Divide both sides by 20:
y = 4

Now that we have the value of y, we can substitute it back into one of the original equations to find x:

x = 8y - 12
x = 8(4) - 12
x = 32 - 12
x = 20

So the solution to this system of equations is x = 20 and y = 4.

Problem 2:
x + 3y = 7
2x - 4y = 24

Similarly, we'll solve one equation for a variable and substitute it into the other equation.

From the first equation, x = 7 - 3y. We can substitute this expression for x in the second equation:

2(7 - 3y) - 4y = 24

Distribute the 2:
14 - 6y - 4y = 24

Combine like terms:
-10y + 14 = 24

Subtract 14 from both sides:
-10y = 10

Divide both sides by -10:
y = -1

Now we can substitute y = -1 into either of the original equations to find x:

x + 3y = 7
x + 3(-1) = 7
x - 3 = 7
x = 10

So the solution to this system of equations is x = 10 and y = -1.

Problem 3:
3a + b = 3
2a - 5b = -15

Let's solve the first equation for one variable and substitute it into the second equation.

From the first equation, b = 3 - 3a. We can substitute this expression for b in the second equation:

2a - 5(3 - 3a) = -15

Distribute the -5:
2a - 15 + 15a = -15

Combine like terms:
17a - 15 = -15

Add 15 to both sides:
17a = 0

Divide both sides by 17:
a = 0

Now that we have the value of a, we can substitute it back into one of the original equations to find b:

3a + b = 3
3(0) + b = 3
0 + b = 3
b = 3

So the solution to this system of equations is a = 0 and b = 3.

Problem 4:
-6 = 3x - 6y
4x = 4 + 5y

In this case, let's solve the second equation for one variable and substitute it into the first equation.

From the second equation, 4x = 4 + 5y, we can solve for x:

x = (4 + 5y) / 4

Now we can substitute this expression for x in the first equation:

-6 = 3((4 + 5y) / 4) - 6y

Multiply both sides by 4 to get rid of the fraction:
-24 = 3(4 + 5y) - 24y

Distribute the 3:
-24 = 12 + 15y - 24y

Combine like terms:
-24 = 12 - 9y

Subtract 12 from both sides:
-36 = -9y

Divide both sides by -9:
4 = y

Now substitute y = 4 into the second equation to solve for x:

4x = 4 + 5(4)
4x = 4 + 20
4x = 24
x = 6

Therefore, the solution to this system of equations is x = 6 and y = 4.

I hope this explanation helps you understand how to solve these problems using the substitution method in Algebra 2. If you have any further questions, feel free to ask!