Respuesta :
Fries are $2.50 and burgers are $4.75.
Our system of equations for this situation would be
[tex] \left \{ {{4b+3f=26.50} \atop {5b+5f=36.25}} \right. [/tex]
We want the coefficients of one of the variables to be the same in order to eliminate one of them; we will make the coefficients of f the same. To do this, multiply the top equation by 5 and the bottom equation by 3:
[tex] \left \{ {{5(4b+3f=26.50)} \atop {3(5b+5f=36.25)}} \right. \\ \\ \left \{ {{20b+15f=132.50} \atop {15b+15f=108.75}} \right[/tex]
We subtract the bottom equation:
[tex] \left \{ {{20b+15f=132.50} \atop {-(15b+15f=108.75)}} \right. \\ \\ 5b = 23.75[/tex]
Divide both sides by 5:
5b/5 = 23.75/5
b = 4.75
Burgers are $4.75.
Substitute this into the first equation:
4(4.75) + 3f = 26.50
19 + 3f = 26.50
Subtract 19 from both sides:
19 + 3f - 19 = 26.50 - 29
3f = 7.50
Divide both sides by 3:
3f/3 = 7.50/3
f = 2.50
Our system of equations for this situation would be
[tex] \left \{ {{4b+3f=26.50} \atop {5b+5f=36.25}} \right. [/tex]
We want the coefficients of one of the variables to be the same in order to eliminate one of them; we will make the coefficients of f the same. To do this, multiply the top equation by 5 and the bottom equation by 3:
[tex] \left \{ {{5(4b+3f=26.50)} \atop {3(5b+5f=36.25)}} \right. \\ \\ \left \{ {{20b+15f=132.50} \atop {15b+15f=108.75}} \right[/tex]
We subtract the bottom equation:
[tex] \left \{ {{20b+15f=132.50} \atop {-(15b+15f=108.75)}} \right. \\ \\ 5b = 23.75[/tex]
Divide both sides by 5:
5b/5 = 23.75/5
b = 4.75
Burgers are $4.75.
Substitute this into the first equation:
4(4.75) + 3f = 26.50
19 + 3f = 26.50
Subtract 19 from both sides:
19 + 3f - 19 = 26.50 - 29
3f = 7.50
Divide both sides by 3:
3f/3 = 7.50/3
f = 2.50
