Two-Step Equations with Fractions - Pre-Algebra

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Solve for :

$\frac{2}{3}$s+\frac{1}{3}$=-$\frac{1}{3}$

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$\frac{2}{3}$s+\frac{1}{3}$=-$\frac{1}{3}$

$\frac{2}{3}$s+\frac{1}{3}$-$\frac{1}{3}$=-$\frac{1}{3}$-$\frac{1}{3}$

$\frac{2}{3}$s=-$\frac{2}{3}$

$$\frac{3}{2}$\cdot \left($\frac{2}{3}$s\right)=\frac{3}{2}$\cdot \left(-$\frac{2}{3}\right)$

Solve for :

$\frac{2}{3}$x-$\frac{1}{4}$=\frac{1}{6}$

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Step 1: Add $\frac{1}{4}$ to both sides:

$\frac{2}{3}$x-$\frac{1}{4}$+\frac{1}{4}$=\frac{1}{6}$+\frac{1}{4}$

$\frac{2}{3}$x+0=\frac{1}{6}$+\frac{1}{4}$

$\frac{2}{3}$x=\frac{1}{6}$+\frac{1}{4}$

Step 2: Add $\frac{1}{6}$ to $\frac{1}{4}$ . Remember that when you add fractions, you must find common denominator. The common denominator for $\frac{1}{4}$ and $\frac{1}{6}$ is . $\frac{1}{6}$ becomes $\frac{2}{12}$ when you multiply both the numerator and the denominator by . Similarly, $\frac{1}{4}$ becomes $\frac{3}{12}$ when you multiply both the numerator and the denominator by .

$\frac{2}{3}$x=\frac{1}{6}$+\frac{1}{4}$

$\frac{2}{3}$x=\frac{2}{12}$+\frac{3}{12}$

$\frac{2}{3}$x=\frac{5}{12}$

Step 3: Multiply both sides of the equation by the reciprocal of $\frac{2}{3}$ :

$\frac{3}{2}$\times $\frac{2}{3}$x=\frac{5}{12}$\times $\frac{3}{2}$

$1x=\frac{15}{24}$$

$x=\frac{15}{24}$$

Step 4: Simplify the fraction by dividing the numerator and the denominator by the Greatest Common Factor (GCF). The GCF of and is :

$x=\frac{15}{24}$=\frac{\frac{15}{3}$}{$\frac{24}{3}$}=\frac{5}{8}$$

$x=\frac{5}{8}$$

Solve for :

$$\frac{1}{2}$x+\frac{3}{4}$=9$

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The goal is to isolate the variable on one side.

$$\frac{1}{2}$x+\frac{3}{4}$=9$

Subtract $\frac{3}{4 }$ from each side of the equation:

$\frac{1}{2}$x+\frac{3}{4}$-$\frac{3}{4}$=9-$\frac{3}{4}$

$\frac{1}{2}$x=8$\frac{1}{4}$

Multiply both sides by :

$x=16$\frac{1}{2}$$

Solve for :

$\frac{2}{3}$x+6$\frac{1}{3}$=10$\frac{1}{3}$

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The goal is to isolate the variable to one side.

$\frac{2}{3}$x+6$\frac{1}{3}$=10$\frac{1}{3}$

First, convert mixed numbers to improper fractions:

$\frac{2}{3}$x+\frac{19}{3}$=\frac{31}{3}$

Subtract $\frac{19}{3}$ from both sides:

$\frac{2}{3}$x+\frac{19}{3}$-$\frac{19}{3}$=\frac{31}{3}$-$\frac{19}{3}$

$\frac{2}{3}$x=\frac{12}{3}$

Multiply each side by the reciprocal of $\frac{2}{3}$ :

$\frac{3}{2}$$\frac{2}{3}$x=\frac{12}{3}$$\frac{3}{2}$

Cross out like terms and multiply:

$\frac{1}{1}$x=\frac{6}{1}$*$\frac{1}{1}$

Solve for $\small x$ :

$$\frac{1}{2}$x+2=4$

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You are trying to isolate the $\small x$ .

To do this you must first subtract both sides by 2 to get

$$\frac{1}{2}$x=2$

This then becomes a one-step problem where you multiply both sides by 2 to get

Solve for x:

$x-$\frac{1}{4}$=\frac{1}{2}$$

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Once you've isolated x, it's important to find the lowest common denominator so that you can add the two fractions you're working with.

Step 1: Isolate x and convert fractions so that they have a common denominator

$x-$\frac{1}{4}$+\frac{1}{4}$=\frac{1}{2}$+\frac{1}{4}$$

$x=\frac{1}{2}$+ $\frac{1}{4}$=\frac{1\cdot 2}{2\cdot 2}$ +\frac{1}{4}$=\frac{2}{4}$+\frac{1}{4}$$

Step 2: solve for x

$x=\frac{2}{4}$+\frac{1}{4}$=\frac{3}{4}$, x=\frac{3}{4}$$

Solve for .

$$\frac{2}{5}$x + $\frac{4}{3}$ = 2$

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First, you want to leave all terms with x on one side and all other terms on the other side. To do this, we can subtract 4/3 from both sides.

We now have

$\frac{2}{5}$x = $\frac{2}{3}$

We can now multiply both sides by the reciprocal of 2/5, which is 5/2, to be able to solve for just x.

$x = $\frac{2}{3}$ * $\frac{5}{2}$ = $\frac{5}{3}$$

Solve for :

$1$\frac{1}{3}$x+7$\frac{1}{3}$=15$\frac{1}{3}$$

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Explanation:

The goal is to isolate the variable to one side.

$1$\frac{1}{3}$x+7$\frac{1}{3}$=15$\frac{1}{3}$$

First, convert the mixed numbers to improper fractions:

$\frac{4}{3}$x+\frac{22}{3}$=\frac{46}{3}$

Subtract $\frac{22}{3}$ from both sides:

$\frac{4}{3}$x+\frac{22}{3}$-$\frac{22}{3}$=\frac{46}{3}$-$\frac{22}{3}$

$\frac{4}{3}$x=\frac{24}{3}$

Multiply each side by the reciprocal of $\frac{3}{4}$ :

$\frac{3}{4}$$\frac{4}{3}$x=\frac{24}{3}$$\frac{3}{4}$

Solve for .

$$\frac{x}$4=\frac{9}$x{}{}$

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Cross multiplication is a short-cut that comes from multiplying by the denominators on both sides of an equation. Broken down, it works like this:

$4\cdot $\frac{x}$4=\frac{9}$x\cdot 4{}{}$

The 4's on the left side of the equation cancel out.

$x=\frac{9\cdot 4}$x{}$

Now, do the same with the denominator on the right side.

$x\cdot x=\frac{9\cdot 4}$x\cdot x{}$

The 's on the right cancel out.

$x\cdot x=9\cdot 4$

This is simply the result of removing the denominators, then multiplying them on the opposite sides, i.e. cross multiplication.

Now, to finish solving for , simplify both sides.

Then take the square root to finish.

$x=$\sqrt{36}$$

$x=\pm 6$

Solve for ""

$$\frac{1}{2}$x -8=4x-1$

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1.) Add 8 to both sides, removing the "". It now reads $$\frac{1}{2}$x=4x+7$

2.) Multiply both sides by 2, removing the $\frac{1}{2}$ . It now reads

3.) Subtract from both sides, removing the "". It now reads

4.) Divide both sides by "", resulting in

Solve for

$$\frac{1}{8}$\ X - 10 = 6$

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$$\frac{1}{8}$ X - 10 = 6$

Add to each side

$$\frac{1}{8}$\ X = 16$

Divide both sides by $\frac{1}{8}$ , or multiply by the reciprocal which is $\frac{8}{1}$

Solve for x: $$\frac{3}{2}$x+\frac{1}{8}$=10$

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To solve, use inverse opperations: do the opposite steps in the opposite order. Order of opperations is usually PEMDAS, with addition and subtraction last, so we'll do addition/subtraction first:

$$\frac{3}{2}$x+\frac{1}{8}$=10$ since it says to add $\frac{1}{8}$ , we will do the opposite and subtract $\frac{1}{8}$ from both sides:

$\frac{3}{2}$x=9 $\frac{7}{8}$ Next we will address multiplication/division. Right now we are multiplying times the fraction $\frac{3}{2}$ . Now we want to multiply both sides times its opposite, the reciprocal $\frac{2}{3}$ :

$x= $\frac{2}{3}$(9 $\frac{7}{8}$)$ The easiest way to do this is to think of the mixed number $9 $\frac{7}{8}$$ as the addition $9+ $\frac{7}{8}$$ and multiply each part times $\frac{2}{3}$ :

$\frac{9}{1}$ $\frac{2}{3}$= 6; $\frac{7}{8}$* $\frac{2}{3}$ = $\frac{7}{12}$

So our answer is $x=6 $\frac{7}{12}$$

Solve the following equation for x.

$\frac{3}{4}$x+\frac{2}{3}$=\frac{1}{6}$x-$\frac{5}{12}$

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To solve equations with fractions, follow the same method as with integers. Collect all the terms with the variable on one side and the terms without variables on the other.

$\frac{3}{4}$x-$\frac{1}{6}$x=-$\frac{5}{12}$-$\frac{2}{3}$

In order to combine the like terms (variables vs. non-variables), the denominators of the fractions must be the same.

$\frac{9}{12}$x-$\frac{2}{12}$x=-$\frac{5}{12}$-$\frac{2}{3}$

$\frac{7}{12}$x=-$\frac{13}{12}$

$x=-$\frac{13}{7}$$

Solve this equation: $6x-$\frac{5}{2}$=\frac{7}{2}$$

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$6x-$\frac{5}{2}$=\frac{7}{2}$$

Move all fractions to one side:

$6x=\frac{7}{2}$+\frac{5}{2}$$

Simplify:

Solve the two-step equation. Find the value of .

$$\frac{x}{4}$ + 222 = 628$

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- 222 - 222

4 4

Check your answer by substituting 1624 back in for x and solving the problem. This time both sides of the equation should match.

Solve for :

$$\frac{1}{8}$x - 14 = 12$

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To solve for the variable, we will need to isolate the variable on one side of the equation and all other contstants on the other side. To do this, apply the opposite operation to manipulate the equation.

First, add to both sides:

$$\frac{1}{8}$x - 14 = 12$

$$\frac{1}{8}$x = 26$

Next, multiply both sides by to solve for :

Solve for :

$$\frac{23}{3y}$ = 3$

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To solve for the variable, we will need to isolate the variable on one side of the equation and all other contstants on the other side. To do this, apply the opposite operation to manipulate the equation.

First, multiply both sides by :

$$\frac{23}{3y}$ = 3$

Next, divide both sides by to solve for :

$y = $\frac{23}{9}$$

$3$\frac{5}{8}$+4$\frac{1}{2}$= ?$

The answer must be a mixed number.

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The first step to adding mixed numbers is to convert them into improper fractions.

$3$\frac{5}{8}$=\frac{[(3\times8)+5]}{8}$ = $\frac{29}{8}$$

$4$\frac{1}{2}$=\frac{[(4\times2)+1]}{2}$=\frac{9}{2}$$

Next, find the least common multiple of 2 and 8 so that both fractions have the same denominator.

$\frac{9}{2}$\times$\frac{4}{4}$=\frac{36}{8}$

Now that the denominators of both fractions are the same, add the fractions.

$\frac{29}{8}$+\frac{36}{8}$=\frac{65}{8}$

Convert the improper fraction back into a mixed number.

65 divided by 8 is 8 remainder 1, or $8$\frac{1}{8}$$ .

Solve for :

$\frac{2}{3}$x-$\frac{4}{3}$=\frac{5}{3}$

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$\frac{2}{3}$x-$\frac{4}{3}$=\frac{5}{3}$

Start by isolating the fraction attached the x variable:

$\frac{2}{3}$x{\color{Red} -$\frac{4}{3}$+\frac{4}{3}$}=\frac{5}{3}$+\frac{4}{3}$

The red terms cancel out.

We add the right side as usual.

$\frac{2}{3}$x=\frac{9}{3}$

Reduce fractions where able and multiply by the reciprocal to isolate x:

${\color{Red} $\frac{3}{2}$$\frac{2}{3}$}x=\frac{9}{3}$$\frac{3}{2}$$

The red terms cancel to 1 and the right is multiplied as usual.

$x=\frac{9}{2}$$

Solve: $\frac{1}{2}$x-$\frac{3}{2}$ = $\frac{5}{2}$

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In order to solve the equation, we have to isolate the variable. We do this by performing the same operation to either side of the equation.

To isolate the unknown variable, first add three halves on both sides of the equation.

$$\frac{1}{2}$x-$\frac{3}{2}$+\left($\frac{3}{2}\right) = $\frac{5}{2}$+\left($\frac{3}{2}\right)$

$\frac{1}{2}$x = $\frac{8}{2}$

Multiply by two on both sides to eliminate the fraction coefficient in front of the variable.

$$\frac{1}{2}$x \times2= $\frac{8}{2}$\times2$