# Thread: setting up a system of equations

1. ## setting up a system of equations

Q: Four large cheeseburgers and two chocolate shakes cost a total of $7.90. Two shakes cost 15 cents more than one cheeseburger. What is the cost of a cheeseburger? A shake? A: I will let x=cost of cheeseburger and y=cost of shake. We know that four large cheesbeburgers and two chocolates shakes costs a total of$7.90; thus, our first equation is 4x+2y=7.90. Now, we also know that two shakes cost 15 cents more than one cheeseburger. This implies, x-2y=0.15. So, the system of equations is,

4x+2y=7.90
x-2y=0.15.

Therefore, x=$1.61 and y=$0.73.

This is not the answer in the back of the book. I am not sure what I am doing wrong.

I got the second question!

Thanks

2. Originally Posted by EPZ

4x+2y=7.90
x-2y=0.15.
try

$
2y-x=0.15
$

3. For the first one: let cheeseburgers = c, let shakes = s

from the question:
4c + 2s = 7.90 (1)
2s = c + 0.15 (2)

putting (2) into (1)

4c + c + 0.15 = 7.90
5c + 0.15 = 7.90
5c = 7.75
c = 7.75 / 5
c = 1.55

you should be able to get both prices from here...

4. Originally Posted by Joel
For the first one: let cheeseburgers = c, let shakes = s

from the question:
4c + 2s = 7.90 (1)
2s = c + 15 (2)
be careful Joel, this system implies \$15 is the difference

Originally Posted by Joel
5c + 15 = 7.90
5c = 7.75
$
7.90-15 = -7.10\neq 7.75
$

5. Q1.

c=cheeseburge, s=shake.

eq1. 4c+2s=7.9
eq2. 2s-0.15=c
Substitute eq2. into eq1.
4(2s-0.15)+2s=7.9
8s-0.6+2s=7.9
10s=8.5
s=0.85
Substitute this into eq2.
2(0.85)-0.15=c
1.7-0.15=c
c=1.55
Check with eq1.
4(1.55)+2(0.85)=7.9

Q2.

c=cashew, p=peanut, m=mixture w=weight

eq1. c=5/w
eq2. p=1.5/w
eq3. 30p+xc=m(30+x)
eq4. m=3/w
Sustitute eq1,2,4. into eq3.
30(1.5/w)+x(5/w)=(3/w)(30+x)
Mutliply all by w (obviously as each weight was per lb).
30(1.5)+x(5)=3(30+x)
45+5x=90+3x
2x=45
x=22.5

We need to ass 22.5 lb's of cashews.