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2 years ago

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In a class there are 15 students. 8 of them like playing soccer, 6 of them like swimming, and 2 like both and swimming and playi

ng soccer. How many students do not like playing soccer and do not like swimming?

1 answer:

xenn [34]2 years ago

5 0

Answer:

6+8-2=12, 15-12=3 answer=3

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Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

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∴ x = 0 is the solution to the inequality -2x < 10.

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-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

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<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

$-2x\:$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)>10\left(-1\right)$

Simplify

$2x>-10$

Divide both sides by 2

$\frac{2x}{2}>\frac{-10}{2}$

$x>-5$

$-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}$

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

$-2x>-6$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)$

Simplify

$2x$

Divide both sides by 2

$\frac{2x}{2}$

$x$

$-6$

Thus, the two intervals:

$\left(-\infty \:,\:3\right)$

$\left(-5,\:\infty \:\right)$

The intersection of these two intervals would be the solution to both inequalities.

$\left(-\infty \:,\:3\right)$ and $\left(-5,\:\infty \:\right)$

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

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3 years ago