Solve One Variable Linear Equations and Inequalities: CCSS.Math.Content.HSA-REI.B.3 - Algebra

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Question

Solve the following inequality for $\uptext{x}$ . Round your answer to the nearest tenth.

$\sqrt{x + 7} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 7} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 7 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 7 -\left( x + 7 \right)> x^{2} + 2 x + 1 -\left( x + 7 \right)$

$0 > x^{2} + x - 6$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 10} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 10} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 10 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 10 -\left( x + 10 \right)> x^{2} + 2 x + 1 -\left( x + 10 \right)$

$0 > x^{2} + x - 9$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

$\frac{3x+4}{4}<\frac{x}{2}$

Which of the following provides the complete solution set for given the above inequality?

Answer

To solve this problem, first cross-multiply the inequality to eliminate the denominators. Note that while this is an inequality, you can safely multiply by both denominators since both are positive so there is no need to consider flipping the direction of the inequality. The result of this step is:

Then you can combine like terms by subtracting from both sides:

Then to isolate the variable term, subtract from both sides:

Finally, divide both sides by to get the variable alone:

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 3} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 3} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 3 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 3 -\left( x + 3 \right)> x^{2} + 2 x + 1 -\left( x + 3 \right)$

$0 > x^{2} + x - 2$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 6} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 6} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 6 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 6 -\left( x + 6 \right)> x^{2} + 2 x + 1 -\left( x + 6 \right)$

$0 > x^{2} + x - 5$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 35} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 35} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 35 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 35 -\left( x + 35 \right)> x^{2} + 2 x + 1 -\left( x + 35 \right)$

$0 > x^{2} + x - 34$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 17} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 17} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 17 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 17 -\left( x + 17 \right)> x^{2} + 2 x + 1 -\left( x + 17 \right)$

$0 > x^{2} + x - 16$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 8} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 8} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 8 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 8 -\left( x + 8 \right)> x^{2} + 2 x + 1 -\left( x + 8 \right)$

$0 > x^{2} + x - 7$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 15} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 15} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 15 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 15 -\left( x + 15 \right)> x^{2} + 2 x + 1 -\left( x + 15 \right)$

$0 > x^{2} + x - 14$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 16} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 16} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 16 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 16 -\left( x + 16 \right)> x^{2} + 2 x + 1 -\left( x + 16 \right)$

$0 > x^{2} + x - 15$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 34} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 34} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 34 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 34 -\left( x + 34 \right)> x^{2} + 2 x + 1 -\left( x + 34 \right)$

$0 > x^{2} + x - 33$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 19} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 19} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 19 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 19 -\left( x + 19 \right)> x^{2} + 2 x + 1 -\left( x + 19 \right)$

$0 > x^{2} + x - 18$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ . Round your answer to the nearest tenth.

$\sqrt{x + 7} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 7} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 7 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 7 -\left( x + 7 \right)> x^{2} + 2 x + 1 -\left( x + 7 \right)$

$0 > x^{2} + x - 6$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 10} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 10} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 10 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 10 -\left( x + 10 \right)> x^{2} + 2 x + 1 -\left( x + 10 \right)$

$0 > x^{2} + x - 9$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 1} > x + 1$

Answer

The first step is to square each side of the inequality.

$\left( \sqrt{x + 1} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 1 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 1 -\left( x + 1 \right)> x^{2} + 2 x + 1 -\left( x + 1 \right)$

$0 > x^{2} + x$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

Compare your answer with the correct one above

Question

$\frac{3x+4}{4}<\frac{x}{2}$

Which of the following provides the complete solution set for given the above inequality?

Answer

To solve this problem, first cross-multiply the inequality to eliminate the denominators. Note that while this is an inequality, you can safely multiply by both denominators since both are positive so there is no need to consider flipping the direction of the inequality. The result of this step is:

Then you can combine like terms by subtracting from both sides:

Then to isolate the variable term, subtract from both sides:

Finally, divide both sides by to get the variable alone:

Compare your answer with the correct one above

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