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

Solve for :

$x\leq11$

Explanation

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

Subtract on both sides.

Divide on both sides.

Solve for .

$x\leq-9$

$x\geq-9$

Explanation

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

Subtract on both sides.

Divide on both sides. Remember to flip the sign.

Solve for .

$x\leq-6$

$x\geq-6$

Explanation

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

Subtract on both sides.

Solve for .

$\left | x+5\right |<8$

Explanation

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

$\left | x+5\right |<8$ We need to set-up two equations since its absolute value.

Subtract on both sides.

Divide on both sides which flips the sign.

Subtract on both sides.

Since we have the 's being either greater than or less than the values, we can combine them to get .

What values of make the statement $\left |5x-9 \right |\geq6$ true?

$x\geq3, x\leq \frac{3}{5}$

$x\geq5,x\leq \frac{1}{5}$

$x\geq4,x\leq -\frac{1}{2}$

$x\geq6,x\leq \frac{1}{3}$

$x\geq15,x\leq \frac{2}{5}$

Explanation

First, solve the inequality $5x-9 \geq6$ :

$5x-9 \geq6$

$5x\geq15$

$x\geq3$

Since we are dealing with absolute value, $5x-9\leq-6$ must also be true; therefore:

$5x-9\leq-6$

$5x\leq3$

$x\leq \frac{3}{5}$

Solve for :

$\left | x+5\right |<2x+16$

Explanation

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

$\left | x+5\right |<2x+16$ We need to set-up two equations since it's absolute value.

Subtract 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 .

$\left | 0+5\right |<2(0)+16$ This is true therefore is a correct answer. Let's next try .

$\left | -8+5\right |<2(-8)+16$ This is not true therefore is not correct.

Final answer is just .

Solve for .

$\left | x+7\right |\leq3x+13$

$-3\leq x$

$-5\leq x$

$-5\leq x\leq -3$

$x\leq -3, x\leq -5$

$3\leq x\leq 5$

Explanation

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

$\left | x+7\right |\leq3x+13$ We need to set-up two equations since it's absolute value.

$x+7\leq 3x+13$ Subtract on both sides.

$-6\leq 2x$ Divide on both sides.

$-3\leq x$

$-(x+7)\leq 3x+13$ Distribute the negative sign to each term in the parenthesis.

$-x-7\leq3x+13$ Add and subtract on both sides.

$-20\leq4x$ Divide on both sides.

$-5\leq x$ We must check each answer. Let's try .

$\left | 0+7\right |<3(0)+13$ This is true therefore $-3\leq x$ is a correct answer. Let's next try .

$\left | -4+7\right |<3(-4)+13$ This is not true therefore $-5\leq x$ is not correct.

Final answer is just $-3\leq x$ .

Solve: $3x+2\le 49$

$x\le \frac{47}{3}$

$x\le 47$

$x\le 17$

$x\le \frac{17}{3}$

$x\ge \frac{47}{3}$

Explanation

To solve $3x+2\le 49$ , isolate .

$3x\le 47$

Divide by three on both sides.

$x\le \frac{47}{3}$

If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?

|w|

_w_2

|w|0.5

w/2

3_w_/2

Explanation

3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.

If $x+1< 4$ and $y-2<-1$ , then which of the following could be the value of ?

Explanation

To solve this problem, add the two equations together:

$x+1<4$

$y-2<-1$

$x+1+y-2<4-1$

$x+y-1<3$

$x+y<4$

The only answer choice that satisfies this equation is 0, because 0 is less than 4.

How to find the solution to an inequality with addition

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