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3 years ago

11

If y=-2 what is the solution for 4(x-3y)<6(x-3y)

1 answer:

Rzqust [24]3 years ago

5 0

Step 1) Plug in

<em>4(-2-3(-2)) < 6(-2-3(-2))</em>

Step 2) Take care of inner parentheses

<em>4(-2+6) < 6(-2+6)</em>

Step 3) Distribute

<em>-8+24 < -12+36</em>

Step 4) Add

16 < 24 (final answer)

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katrin2010 [14]

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In sentences why can’t x^2+2x+12 cannot be factored

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$x^2+2x+12$

This cannot be factored because you can'y take a number out.

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Step-by-step explanation:

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3 years ago

Which of the following ordered pairs is not a solution of the system of inequalities? y > x2 – 4x – 5 y < –x2 – 5x + 6 Que

horrorfan [7]

Answer:

a) (x, y) = (1, 7)

Step-by-step explanation:

Let be the following system of inequalties:

$y > x^{2}-4\cdot x -5$

$y < -x^{2}-5\cdot x +6$

We can find the right option by evaluating each option in the system of inequalities:

a) (x, y) = (1, 7)

$7 > 1^{2}-4\cdot (1) -5$

$7 < -1^{2}-5\cdot (1) +6$

Then,

$7>-8$ (TRUE)

$7$ (FALSE)

(1, 7) is not a solution of the system of inequalities.

b) (x, y) = (1, -7)

$-7 > 1^{2}-4\cdot (1) -5$

$-7 < -1^{2}-5\cdot (1) +6$

Then,

$-7 > - 8$ (TRUE)

$-7< 0$ (TRUE)

(1, -7) is a solution of the system of inequalities.

c) (x, y) = (1, -5)

$-5 > 1^{2}-4\cdot (1) -5$

$-5 < -1^{2}-5\cdot (1) +6$

Then,

$-5 > - 8$ (TRUE)

$-5< 0$ (TRUE)

(1, -5) is a solution of the system of inequalities.

d) (x, y) = (-1, 6)

$6 > (-1)^{2}-4\cdot (-1) -5$

$6$

Then,

$6>0$ (TRUE)

$6 < 12$ (TRUE)

(-1, 6) is a solution of the system of inequalties.

Therefore, we conclude that correct answer is A.

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Step-by-step explanation:

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