SOLUTION: Larry spent 1/2 of his money on a camera and 1/8 on a radio. The camera cost $120 more that the radio. How much money did he start with?

Question 821817: Larry spent 1/2 of his money on a camera and 1/8 on a radio.
The camera cost $120 more that the radio.
How much money did he start with?
Found 2 solutions by TimothyLamb, richwmiller:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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x = starting cash
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the camera:
z = (1/2)x
---
the radio:
y = (1/8)x
---
given: the camera is $120 more than the radio
z = y + 120
---
y + 120 = (1/2)x
y = (1/8)x
---
put the system of linear equations into standard form
---
0.5x - y = 120
0.125x - y = 0
---
copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
---
x= 320
y= 40
---
answer:
starting cash = $320
camera = $160
radio = $40
---
Solve and graph linear equations:
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---
Solve quadratic equations, quadratic formula:
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---
Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
c=1/2x,
r=1/8x,
c=r+180
c = 240, r = 60, x = 480
he started with $480
he spent $60 on the radio and $240 on the camera.

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