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adelina 88 [10]
2 years ago
5

Solve by any method (graphing, substitution or linear combination)

Mathematics
1 answer:
Lostsunrise [7]2 years ago
8 0

Answer:

b) (5, -3)

Step-by-step explanation:

The given equations are;

y=x-8...(1)

2x+3y=1...(2)

I prefer using substitution because of the first equation.

Put equation (1) into equation (2) to obtain;

2x+3(x-8)=1

We expand the parenthesis to obtain;

2x+3x-24=1

Group similar terms to get;

2x+3x=1+24

Simplify

5x=25

Divide through by 5;

x=\frac{25}{5}

\Rightarrow x=5

Put x=5 into equation (1) to get;

y=5-8

\Rightarrow y=-3

The solution is (5,-3).

The correct answer is B

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<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

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