# Solutions to equations

• May 7th 2007, 04:00 PM
Ash
Solutions to equations
Can you show me how to go about solving these problems? I'm suppose to translate the given information into a system of equations and solve the system to determine the two numbers.

#1. One number exceeds a second number by 8. Three-fourths of the larger number is 3 more than the smaller number.

With the problem #2, I want to make sure I'm doing it right.
I'm suppose to determine whether each ordered pair is a solution of the given system of linear equations.

#2 (-0.5,0)

2x+3y=3
8x=12y-4

When I plug (-0.5,0) into the equation I get

2(-0.5) + 3(0)=3
-3 = 3 ,

8(-0.5) = 12(0)-4
- 4.0 = 4
-4 = -4 ,

So would (-0.5,0) not be a solution to both equations since one 3 is negative and the other is a positive 3? I know it would of been a solution if it was -4 = -4 and the other was 3 = 3, but I'm not sure since one three is negative.
• May 7th 2007, 04:09 PM
alinailiescu
Quote:

Originally Posted by Ash
Can you show me how to go about solving these problems? I'm suppose to translate the given information into a system of equations and solve the system to determine the two numbers.

#1. One number exceeds a second number by 8. Three-fourths of the larger number is 3 more than the smaller number.

With the problem #2, I want to make sure I'm doing it right.
I'm suppose to determine whether each ordered pair is a solution of the given system of linear equations.

#2 (-0.5,0)

2x+3y=3
8x=12y-4

When I plug (-0.5,0) into the equation I get

2(-0.5) + 3(0)=3
-3 = 3 ,

8(-0.5) = 12(0)-4
- 4.0 = 4
-4 = -4 ,

So would (-0.5,0) not be a solution to both equations since one 3 is negative and the other is a positive 3? I know it would of been a solution if it was -4 = -4 and the other was 3 = 3, but I'm not sure since one three is negative.

#1 Let x and y be the two numbers, assuming that x is the largest.
Since"One number exceeds a second number by 8. ",
x=y+8
the second sentence provides the second equation:"Three-fourths of the larger number is 3 more than the smaller number."
3/4x=y+3
solving the system you have the solution: x=20, y=12
• May 7th 2007, 04:12 PM
alinailiescu
#2 You are right, the pair is not a solution .
Pay attention when evaluating expresion. respect the order of operation. Multiplication first, then addition.
2(-0.5) + 3(0)=3
-1+0=3
-1=3 is not true, so right away you can say that the pair is not a solution.
• May 7th 2007, 05:01 PM
Ash
Quote:

Originally Posted by alinailiescu
#1 Let x and y be the two numbers, assuming that x is the largest.
Since"One number exceeds a second number by 8. ",
x=y+8
the second sentence provides the second equation:"Three-fourths of the larger number is 3 more than the smaller number."
3/4x=y+3
solving the system you have the solution: x=20, y=12

Thanks for your help, I now see where I went wrong on the second problem, but can you show me the steps in how you came up with x= 20 and y=12 ?
• May 7th 2007, 06:17 PM
alinailiescu
You have the system:
x=y+8
3/4x=y+3
using substitution method, by writting y+8 (from the first equation) insted of x into the second we have:
3/4(y+8)=y+3
3/4y+6=y+3
6-3=y-3/4y
3=1/4y
12=y,
using this value into the first equation we have x=12+8so, x=20