How to find the solution to an inequality with subtraction - SAT Math

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Question

If $\frac{a}{5}+5> 6$ , which of the following MUST be true?

I. $a> 2$

II. $a> 10$

III. $a< 6$

Answer

Subtract 5 from both sides of the inequality:

$\frac{a}{5}> 1$

Multiply both sides by 5:

$a> 5$

Therefore only I must be true.

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Question

Which of the following is equivalent to $\left | x-3 \right |<2$ ?

Answer

Solve for both x – 3 < 2 and –(x – 3) < 2.

x – 3 < 2 and –x + 3 < 2

x < 2 + 3 and –x < 2 – 3

x < 5 and –x < –1

x < 5 and x > 1

The results are x < 5 and x > 1.

Combine the two inequalities to get 1 < x < 5

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Question

A factory packs cereal boxes. Before sealing each box, a machine weighs it to ensure that it is no lighter than 356 grams and no heavier than 364 grams. If the box holds grams of cereal, which inequality represents all allowable values of ___?

Answer

The median weight of a box of cereal is 360 grams. This should be an allowable value of w. Substituting 360 for w into each answer choice, the only true results are:

$\ |w -360| \leq 4 \ |360 -360| \leq 4 \0 \leq 4$

and:

$\|w + 360| \geq 4 \ |360 + 360| \geq 4\ 720 \geq4$

Notice that any positive value for w satisfies the second inequality above. Since w must be between 356 and 364, the first inequality above is the only reasonable choice.

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Question

|12x + 3y| < 15

What is the range of values for y, expressed in terms of x?

Answer

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

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Question

|4x + 14| > 30

What is a possible valid value of x?

Answer

This inequality could be rewritten as:

4x + 14 > 30 OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

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Question

Given the inequality, |2_x_ – 2| > 20,

what is a possible value for x?

Answer

For this problem, we must take into account the absolute value.

First, we solve for 2_x_ – 2 > 20. But we must also solve for 2_x_ – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).

First step:

2_x_ – 2 > 20

2_x_ > 22

x > 11

Second step:

2_x_ – 2 < –20

2_x_ < –18

x < –9

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

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Question

$\dpi{100} \small -2y+7>-7+y$

Given the inequality above, which of the following MUST be true?

Answer

$\dpi{100} \small -2y+7>-7+y$ Subtract from both sides:

Subtract 7 from both sides:

Divide both sides by $\dpi{100} \small -3$ :

$\frac{-3y}{-3}>\frac{-14}{-3}$

Remember to switch the inequality when dividing by a negative number:

$y<\frac{14}{3}$

Since $\dpi{100} \small y<\frac{14}{3}$ is not an answer, we must find an answer that, at the very least, does not contradict the fact that is less than (approximately) 4.67. Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that is less than 5.

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Question

The cost, in cents, of manufacturing $\dpi{100} \small x$ pencils is $\dpi{100} \small 1200+20x$ , where 1200 is the number of cents required to run the factory regardless of the number of pencils made, and 20 represents the per-unit cost, in cents, of making each pencil. The pencils sell for 50 cents each. What number of pencils would need to be sold so that the revenue received is at least equal to the manufacturing cost?

Answer

If each pencil sells at 50 cents, $\dpi{100} \small x$ pencils will sell at $\dpi{100} \small 50x$ . The smallest value of $\dpi{100} \small x$ such that

$\dpi{100} \small 50x\geq 1200+20x$

$\dpi{100} \small x\geq 40$

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Question

Solve for $x$ .

$-2x+5\leq 10$

Answer

Move +5 using subtraction rule which will give you $-2x\leq 5$ .

Divide both sides by 2 (using division rule) and you will get $-x\leq \frac{5}{2}$ which is the same as $x\geq \frac{5}{2}$

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Question

If 2 more than is a negative integer and if 5 more than is a positive integer, which of the following could be the value of ?

Answer

and , so and . The only integers between and are and .

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Question

Given $\left | 7x+9\right |>30$ , what is a possible value of ?

Answer

In order to find the range of possible values for $\left | 7x+9\right |>30$ , we must first consider that the absolute value applied to this inequality results in two separate equations, both of which we must solve:

and .

Starting with the first inequality:

Then, our second inequality tells us that

$x<-\frac{39}{7}(or -5\frac{4}{7})$

Therefore, is the correct answer, as it is the only value above for which (NOT greater than or equal to) or $x<-\frac{39}{7}$ (NOT less than or equal to).

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Question

Solve for :

Answer

The correct method to solve this problem is to substract 5 from both sides. This gives .

Then divide both sides by negative 3. When dividing by a negative it is important to remember to change the inequality sign. In this case the sign goes from a less than to a greater than sign.

This gives the answer .

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Question

Solve for .

$4x-8\geq 28$

Answer

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

$4x-8\geq 28$ Add on both sides.

$4x\geq 36$ Divide on both sides.

$x\geq 9$

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Question

Solve for .

Answer

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

Add on both sides.

Divide on both sides. Remember to flip the sign.

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Question

Solve for .

Answer

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

Add and subtract on both sides.

Divide on both sides.

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Question

Solve for .

$\left | 6x-45\right |\leq 75$

Answer

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

$\left | 6x-45\right |\leq 75$ We need to set-up two equations since its absolute value.

$6x-45\leq 75$ Add on both sides.

$6x\leq 120$ Divide on both sides.

$x\leq 20$

$-(6x-45)\leq 75$ Divide on both sides which flips the sign.

$6x-45\geq -75$ Add on both sides.

$6x\geq -30$ Divide on both sides.

$x\geq -5$

Since we have the 's being either greater than or less than the values, we can combine them to get $-5\leq x\leq 20$ .

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Question

Solve for .

$\left | x-19\right |<2x-26$

Answer

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

$\left | x-19\right |<2x-26$ We need to set-up two equations since it's absolute value.

Subtract and add on both sides.

Distribute the negative sign to each term in the parenthesis.

Add , on both sides.

Divide on both sides.

We must check each answer. Let's try .

$\left | 10-19\right |<2(10)-26$ This is not true therefore is not correct. Let's try .

$\left | 20-19\right |<2(20)-26$ This true so therefore is correct.

Final answer is .

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Question

Solve for .

$\left | 5x-75\right |>45-x$

Answer

We want to isolate the variable on one side and numbers on another side. Treat like a normal equation.

$\left | 5x-75\right |>45-x$ We need to set-up two equations since it's absolute value.

Add on both sides.

Divide on both sides.

Distribute the negative sign to each term in the parenthesis.

Add and subtract on both sides.

Divide on both sides.

We must check each answer. Let's try .

$\left | 10(5)-75\right |>45-10$ This is not true therefore is not correct. Let's try .

$\left | 5(30)-75\right |>45-30$ This true so therefore is correct.

Final answer is .

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