For this case, the first thing we must do is define variables.
We have then:d
x: cost of each small pizza
y: cost of each liter of soda
z: cost of each salad.
We write the equation that models the problem:
[tex] 2x + y + z = 14 [/tex]
[tex] x + y + 3z = 15 [/tex]
[tex] 3x + y + 2z = 22 [/tex]
From equation 3 we clear y:
[tex] y = 22 - 3x -2z [/tex]
Substituting in equation 2 we have:
[tex] x + (22 - 3x -2z) + 3z = 15 [/tex]
Rewriting:
[tex] -2x + z = -7 [/tex]
From here, we clear the value of z:
[tex] z = -7 + 2x [/tex]
Then, we substitute the value of z and y in equation 1:
[tex] 2x + (22 - 3x -2 (-7 + 2x)) + (-7 + 2x) = 14 [/tex]
From here, we clear x:
[tex] 2x + (22 - 3x + 14 - 4x) + (-7 + 2x) = 14 [/tex]
[tex] 2x + (36 - 7x) + (-7 + 2x) = 14 [/tex]
[tex] -3x + 29 = 14 [/tex]
[tex] -3x = 14-29 [/tex]
[tex] -3x = -15 [/tex]
[tex] x = 5 [/tex]
Then, the value of z is:
[tex] z = -7 + 2x [/tex]
[tex] z = -7 + 2 (5) [/tex]
[tex] z = -7 + 10 [/tex]
[tex] z = 3 [/tex]
Finally, the value of y is:
[tex] y = 22 - 3x -2z [/tex]
[tex] y = 22 - 3 (5) -2 (3) [/tex]
[tex] y = 22 - 15 - 6 [/tex]
[tex] y = 1 [/tex]
Answer:
$ 5: cost of each small pizza
1 $: cost of each liter of soda
3 $: cost of each salad.
