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2 years ago

Assume you wanted to save $2,000 to buy a used car 18 months from now. How

Mathematics

1 answer:

Scrat [10]2 years ago

3 0

Answer:

27 dollars a week

Step-by-step explanation:

Well there’s 72 weeks in 18 months so 2,000 divided 72 equals 27.

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Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

$x = \frac{2}{3}$

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

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