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How many erasers did Brandon and Perry have in the end?

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$c+b+p=103\\ b:p=4:5\\ \dfrac{c}{2}+\dfrac{b}{2}+p=79\\ p=\dfrac{5b}{4}\\ c+b+\dfrac{5b}{4}=103\\ \dfrac{c}{2}+\dfrac{b}{2}+\dfrac{5b}{4}=79$

$c+b+\dfrac{5b}{2}=158\\ 158-b-\dfrac{5b}{2}=103-b-\dfrac{5b}{4}\\ 55=\dfrac{5b}{2}-\dfrac{5b}{4}\\ 55=\dfrac{5b}{4}\\ \color{blue}b=44\\ Brandon\ has\ 44\ erasers\ at\ the\ beginning.\\ b:p=4:5\\ p=\dfrac{5b}{4}=\dfrac{5\cdot 44}{4}\\ \color{blue}p=55\\Perry\ has\ 55\ erasers\ at\ the\ beginning.$

$c=103-44-55\\ c=4\\ Caden\ has\ 4\ erasers\ at\ the\ beginning.\\ Brandon\ and\ Caden\ gave\ up\ half\ of\ their\ erasers.\\b-\dfrac{b}{2}+p= 44-\dfrac{44}{2}+55=\color{blue}77$

77 erasers did Brandon and Perry have in the end.

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C + B + P==103, 

B/P ==4/5, 

[C - 1/2C + B - 1/2B + P]==79, solve for B, C, P

Use substitution to get:

B==44 erasers - what Brandon started with

C ==4 erasers - what Caden started with

P ==55 erasers - what Perry started with.

44 - (44/2) ==22 erasers - Brandon had in the end.

55 - 0 ==55 erasers - what Perry had in the end.

22 + 55==77 total erasers of Brandon and Perry at the end.

4 - (4/2) ==2 erasers - what Caden had in the end.

Check: 22 + 55 + 2 ==79 total of all erasers at the end.