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3 years ago

Can you solve this for me

Mathematics

1 answer:

jenyasd209 [6]3 years ago

4 0

The number in the blank box should be 44 because it said that "3 times x and 4 times y is 44".

You need to multiply the first equation by 3:

3x + 3y = 39.

Then, subtract the first equation from the second equation:

3x - 3x = 0

4y - 3y = y

44 - 39 = 5

Your new equation should look like this: y = 5

Now, substitute y = 5 into the first original equation:

x + 5 = 13

x = 8

Ans. (x, y) = (8, 5)

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3 years ago

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3 years ago

-336x2y3<br> Reduce<br> to lowest terms

Alecsey [184]

Answer:

$-\frac{42y}{x}$

Step-by-step explanation:

Apply the fraction rule: $\frac{-a}{b}=-\frac{a}{b}$

$=-\frac{336x^2y^3}{8x^3y^2}$

Divide the numbers: $\frac{336}{8} =42$

$=\frac{42x^2y^3}{x^3y^2}$

Apply the exponent rule: $\frac{x^a}{x^b}=\frac{1}{x^{b-a}}$

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

$=\frac{42y^3}{y^2x^{3-2}}$

Subtract the numbers: $\:3-2=1$

$=\frac{42y^3}{xy^2}$

Apply the exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$\frac{y^3}{y^2}=y^{3-2}$

$=\frac{42y^{3-2}}{x}$

Subtract the numbers: $3-2=1$

$=-\frac{42y}{x}$

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3 years ago