SOLUTION: I'm struggling on 2 algebra problems I have. If you could please help me I would greatly appreaciate it. 1) The sum of two numbers is 43. One number plus three times the other

Question 158188: I'm struggling on 2 algebra problems I have. If you could please help me I would greatly appreaciate it.
1) The sum of two numbers is 43. One number plus three times the other number is 65. What are the numbers?
x + x = 43
x + 3x = 65

2) The difference between two numbers is 55. Four times the smaller number plus five times the larger number is 176. What are the numbers?
z = smaller number
y = larger number
x - x = 55
4z + 5y = 76

Found 3 solutions by Alan3354, nerdybill, Edwin McCravy:
Answer by Alan3354(21605)   (Show Source):
You can put this solution on YOUR website!
1) The sum of two numbers is 43. One number plus three times the other number is 65. What are the numbers?
x + x = 43 ??
x + 3x = 65 ??
------------
x + y = 43
x + 3y = 65
---------------
x = 32
y = 11

2) The difference between two numbers is 55. Four times the smaller number plus five times the larger number is 176. What are the numbers?
z = smaller number
y = larger number
x - x = 55
4z + 5y = 76
------------
y - z = 55
4z + 5y = 176
-----
y = 55 + z
Sub for y in 2nd eqn
4z + 5*(55+z) = 176
4z + 275 + 5z = 176
9z = -99
z = -11
y = 44

Answer by nerdybill(5410)   (Show Source):
You can put this solution on YOUR website!
1) The sum of two numbers is 43. One number plus three times the other number is 65. What are the numbers?
x + x = 43
x + 3x = 65
You can't use the SAME variable to represent the two different numbers! Use another variable:
Let x = one number
and y = second number
then
x + y = 43 (equation 1)
x + 3y = 65 (equation 2)
.
solve equation 1 for y:
x + y = 43
y = 43 - x
.
Substitute the above into equation 2 and solve for x:
x + 3y = 65
x + 3(43 - x) = 65
x + 129 - 3x = 65
129 - 2x = 65
129 = 2x+65
64 = 2x
32 = x
.
Substitute the above into equation 1 and solve for y:
x + y = 43
32 + y = 43
y = 11
.
Solution: the numbers are 11 and 32
.
2) The difference between two numbers is 55. Four times the smaller number plus five times the larger number is 176. What are the numbers?
z = smaller number
y = larger number
x - x = 55 <<<<--how did you get x? should be y-z=55
4z + 5y = 76 <<<<--should be 176
.
Let's start over:
z = smaller number
y = larger number
y - z = 55
4z + 5y = 176
.
Solve equation 1 for y:
y - z = 55
y = z+55
.
Substitute the above into equation 2 and solve for z:
4z + 5y = 176
4z + 5(z+55) = 176
4z + 5z + 275 = 176
4z + 5z + 275 = 176
9z + 275 = 176
9z = -99
z = -11
.
Substitute the above into equation 1 and solve for y:
y - z = 55
y - (-11) = 55
y+11 = 55
y = 44
.
Solution: the two numbers are -11 and 44

Answer by Edwin McCravy(6937)   (Show Source):
You can put this solution on YOUR website!
Edwin's explanation of your error:
I'm struggling on 2 algebra problems I have. If you could please help me I would greatly appreaciate it.
1) The sum of two numbers is 43. One number plus three times the other number is 65. What are the numbers?
x + x = 43
x + 3x = 65

I'm afraid that's wrong. Your mistake is trying to use the SAME letter x for two DIFFERENT numbers. You must use DIFFERENT letters for DIFFERENT numbers. First number = Second number = Solve the first equation for Substitute for is the second equation, then solve for Substitute for in ------------------------

2) The difference between two numbers is 55. Four times the smaller number plus five times the larger number is 176. What are the numbers?
z = smaller number
y = larger number
x - x = 55
4z + 5y = 176

Hmm! Now your equation is OK, but your first equation should not have involved at all, but and only. And again, on top of that, your mistake was using the SAME letter for two DIFFERENT quantities. You must use DIFFERENT letters for DIFFERENT quantities, never the SAME letter. Your first equation should have been So you have this system: Solve the first equation for Substitute for is the second equation, then solve for Substitute for in Edwin AnlytcPhil@aol.com