you spent 1/2 of your money on a camera and 18 on a radio. the camera cost $120 more than the radio. How much money did you start with?
Do you mean "1/8" on a radio?
c = 240, r = 60, x = 480
he started with $480
he spent $60 on the radio and $240 on the camera.
x = starting cash
the camera: z = (1/2)x
the radio:y = (1/8)x
the camera is $120 more than the radioz = 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
answer:
starting cash = $320
camera = $160
radio = $40
To find out how much money you started with, let's break down the information given:
Let's assume the money you started with is represented by the variable 'M'.
According to the problem, you spent 1/2 of your money on a camera and $18 on a radio. So we can set up the equation:
(1/2)M + $18 = M
Now, we are also given that the camera cost $120 more than the radio. Therefore, we can set up another equation:
(1/2)M - $18 = $120
Now, we have a system of two equations with two variables. We can solve this system to find the value of 'M' which represents the money you started with.
Here is how we can solve this system of equations:
Step 1: Simplify the first equation:
(1/2)M + $18 = M
Multiply both sides of the equation by 2 to eliminate the fraction:
M + $36 = 2M
Subtract 'M' from both sides of the equation:
$36 = M -----> This gives us the value of 'M', which represents the money you started with.
Therefore, you started with $36.