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half of Robert's piece of wire is equal to 2 thirds of Maria's wire. The total lenght of their wire is 10 feet. How much longer is Robert's wire than Maria's?

Sagot :

CatieSlotz
You have a system of equations here. R=Robert's piece of wire M=Maria's piece of wire

R/2=2M/3
and
R+M=10

With some basic algebra, you can make that first equation into
R=4M/3
then use substitution to get
(4M/3)+M=10
Then solve for M
(4M/3)+(3M/3)=10
7M/3=10
7M=30
M=30/7

so Maria's wire is 30/7 feet long. With a little more substitution, we can get:
R+(30/7)=10
R=10-(30/7)
R=(70/7)-(30/7)
R=40/7 feet long

So your final answer would be that Robert's wire is 10/7 feet longer than Maria's (which you could always put into decimal form)
Evadin2019

Robert's wire is [tex]1\frac{3}{7}[/tex] longer than Maria's wire

Further Explanation

Let's say

Robert's wire = x

Maria's wire = y

From the problem given we know that:

[tex]\boxed {\frac{1}{2} x = \frac{2}{3}y }\\ \boxed {x + y = 10 }[/tex]

[tex]\boxed {x = 10-y }[/tex]

So we get two equations:

[tex]\boxed {\frac{1}{2} x = \frac{2}{3}y }\\\boxed {\frac{1}{2} (10-y) = \frac{2}{3}y }  \\\boxed {5 - \frac{1}{2}y = \frac{2}{3} y }\\\boxed {5 = \frac{2}{3}y + \frac{1}{2}y }\\\boxed {5 = \frac{7}{6} y }\\\boxed { y= \frac{30}{7} }[/tex]

Let's subsitute the variable y into one of the equation above:

[tex]\boxed {x + y = 10 }\\\boxed {x + \frac{30}{7} = 10 }\\\boxed { x = 10 -\frac{30}{7} }\\\boxed { x = \frac{70-30}{7} }\\\boxed { x = \frac{40}{7} }[/tex]

Robert's wire (x) is [tex]\frac{40}{7}[/tex] feet long and Maria's wire is [tex]\frac{30}{7}[/tex] long.

The difference between Robert's and Maria's is

[tex]\boxed {= \frac{40}{7} - \frac{30}{7} }\\\boxed {= \frac{10}{7} }\\\boxed {= 1\frac{3}{7} }[/tex]

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Keywords: multi-step problem, fraction, part to whole relationship, mixed fraction, additional fraction, subtraction fraction from a whole number

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