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!