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2 years ago

3/4 is added to x, the result is 9/8. What is the value of x?

Mathematics

1 answer:

KIM [24]2 years ago

4 0

Answer:

x=3/8

Step-by-step explanation:

turn 3/4 into a fraction that has a denominator of 8 3/4x2= 6/8

To get it to 9/8 you add x which is 3/8

so 3/8+6/8 is 9/8

x=3/8

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Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.

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Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

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Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

Solving the expression

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

Solving the expression

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

Solving the expression

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

Solving the expression

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

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