1. ## system of equations

ok, here's the question...

Bechorath bought 3 times as many toy cars as game cartridges. He spent $1615 altogether. A game cartridge costs$15 more than a toy car. The total cost of the toy cars was $425 more than the total cost of game cartridges. Find the cost of a game cartridge. I would be gald if u help me..... 2. Originally Posted by BlAcK-lIsTeD ok, here's the question... Bechorath bought 3 times as many toy cars as game cartridges. He spent$1615 altogether. A game cartridge costs $15 more than a toy car. The total cost of the toy cars was$425 more than the total cost of game cartridges. Find the cost of a game cartridge.

I would be gald if u help me.....
Let the cost of a game cartridge be x.
Let the cost of a toy car be y.

Then the total cost is
mx + ny = 1615
where m is the number of cartridges bought and n is the number of toy cars bought. And we know that n = 3m, so

mx + 3my = 1615

Now, the cost of a game cartridge is $15 more than the cost of a toy car, so x = y + 15: m(y + 15) + 3my = 1615 The total cost of the toy cars, 3my, is$425 more than the total cost of the of the game cartridges, m(y + 15), so:
3my = m(y + 15) + 425

So we now have two equations and two unknowns:
m(y + 15) + 3my = 1615 ==> 4my + 15m = 1615
3my = m(y + 15) + 425 ==> 2my - 15m = 425

Multiply the bottom equation by 2:
4my + 15m = 1615
4my - 30m = 850

Now subtract the bottom equation from the top:
(4my + 15m) - (4my - 30m) = 1615 - 850

45m = 765

m = 17.

Now put m = 17 into one of our two equations. I'll pick on the bottom one again:
2my - 15m = 425

2(17)y - 15(17) = 425

34y - 255 = 425

34y = 680

y = 20

Thus
x = y + 15 = 20 + 15 = 35

So a game cartridge is \$35.

-Dan

3. ## thank u but....

thanks for helping me but can u explain it in a easier way? i m really bad in maths....thanks a bunch!