One bar of candy a and two bars of candy b have 781 calories. two bars of candy a and one bar of candy b contain 794 calories. find the calories of each candy bar

A+2B=781
2A+B=794
elmimante one
times firs equation by -2 and add to 2nd

-2A-4B=-1562
2A+b=794 +
0A-3B=-768

-3B=-768
divide both sides by -3
B=256

subsitute

2A+B=794
2A+256=794
minus 256 both sides
2A=538
divide 2 both sides
A=269

A has 269 calories
B has 256 calories

Let x = candy A
let y = candy B

x + 2y = 781
2x + y = 794

2x + y = 794
-2x -2x
y = 794 - 2x

x + 2 (794 - 2x) = 781 ▪distributive property
x + 1588 - 4x = 781 ▪ combine like terms
-3x + 1588 =781
- 1588 -1588

-3x = - 807
÷ (-3) ÷ (-3)
x = 269

Candy A contains 269 calories

▪ Plug x into one of the original equations

x + 2y = 781
269 + 2y = 781
-269 -269

2y = 512
÷2 ÷2

y = 256

Candy B contains 256 calories

The point (5, 10) will work for the rule y=x+5.y= 10 x= 510=5+510=10therefore the ordered pair (5, 10) works for the rule