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3 years ago

Which ordered pairs make the equation true?

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

6 0

Answer:

(-3, 1) and (2, -1)

Step-by-step explanation:

We'll just substitute in these coordinates.

First Pair:

2(-3) + 5(1) = -1

-6 + 5 = -1

-1 = -1

This works!

Second Pair:

2(5) + 5(-3) = -1

10 - 15 = -1

-5 ≠ -1

This does not work.

Third Pair:

2(2) + 5(-1) = -1

4 - 5 = -1

-1 = -1

This also works!

Fourth Pair:

2(-4) + 5(2) = -1

-8 + 10 = -1

2 ≠ -1

This does not work.

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Step-by-step explanation:

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3 years ago

What is the solution of the following system of equations ? y- 2x and= 6 , y - 4x =0

Sveta_85 [38]

Answer:

x=3 and y=12

Step-by-step explanation:

$y-2x=6$ equation:1

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3 years ago

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2 years ago

Read 2 more answers

Rationalize the denominator and simplify. StartFraction 5 minus StartRoot 3 EndRoot Over 4 plus 2 StartRoot 3 EndRoot EndFractio

nasty-shy [4]

Answer:

$\frac{13-7\sqrt{3}}{2}$

Step-by-step explanation:

We need to rationalize the denominator of $= \frac{5-\sqrt{3}}{4+2\sqrt{3}}$ . For rationalizing we multiply the equation by $\frac{4-2\sqrt{3}}{4-2\sqrt{3}}$

So, solving

$= \frac{5-\sqrt{3}}{4+2\sqrt{3}}*\frac{4-2\sqrt{3}}{4-2\sqrt{3}} \\=\frac{(5-\sqrt{3})(4-2\sqrt{3})}{4+2\sqrt{3}*4-2\sqrt{3}}\\=\frac{(5-\sqrt{3})(4-2\sqrt{3})}{(4)^2-(2\sqrt{3})^2}\\= \frac{5(4-2\sqrt{3})-\sqrt{3}(4-2\sqrt{3})}{16-(4*3)}\\=\frac{20-10\sqrt{3}-4\sqrt{3}+2*3}{16-12}\\=\frac{20+6-14\sqrt{3}}{4}\\=\frac{26-14\sqrt{3}}{4}\\= \frac{2(13-7\sqrt{3})}{4}\\=\frac{13-7\sqrt{3}}{2}$

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3 years ago