Algebra fraction solver: an algebra solver with fractions

Algebra solver with fractions: the method it follows

An equation with fractions looks harder than it is, because there is one idea behind every case: multiply both sides by a number that makes every denominator disappear. That number is the least common denominator. For x/2 + x/3 = 5 it is 6, and multiplying by 6 turns the equation into 3x + 2x = 30, an ordinary linear equation.

This algebra solver with fractions writes each of those moves on its own line with a one-sentence reason. It handles two kinds of problem. When the unknown is in a denominator, the equation is only defined for some values of x, and those values are written down first.

Why denominators with x need a restriction

A fraction is undefined when its denominator is zero. In 1/x + 1/(x + 1) = 5/6 the value x = 0 would put zero under the first fraction and x = −1 would put zero under the second, so neither can be a solution, whatever the algebra later produces. In the last example, (x + 1)/(x − 2) = 3/(x − 2), the algebra gives x = 2, but 2 is exactly the value that makes the denominator zero. The equation has no solution, and the solver says so instead of reporting 2.

How to use it

1. Type each fraction with a slash and put the top or bottom in brackets when it has more than one term, for example (2x - 1)/3.
2. Add "for x" if the equation has more than one letter.
3. Read the first step. With x in a denominator it lists the values x cannot be, and those are ruled out for the rest of the solution.
4. Check the last line against the original equation, and compare the answer with the list of values x cannot be.

Algebra fraction solver: worked examples with checks

Each example is the solver's own output for the equation shown. We worked every one out by hand as well, and the answers match. Use the link under an example to run it on the board above.

1. Solve x/2 + x/3 = 5 for x

1. Find the least common denominatorLCD(2, 3) = 6Determine the least common multiple of the denominators 2 and 3 to eliminate the fractions.
2. Multiply both sides by 66 * (x/2 + x/3) = 6 * 5Multiply every term on both sides by the LCD to clear the denominators.
3. Distribute and simplify3x + 2x = 30Apply the distributive property: 6*(x/2) simplifies to 3x, and 6*(x/3) simplifies to 2x.
4. Combine like terms5x = 30Add the coefficients of x together: 3 + 2 = 5.
5. Divide both sides by 5x = 30 / 5 = 6Divide by 5 to isolate the variable x.

Answer: x = 6

Check: Substitute x = 6 back into the original equation: 6/2 + 6/3 = 3 + 2 = 5, which matches the right-hand side.

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3. Solve 1/x + 1/(x+1) = 5/6 for x

1. Identify the domain restrictionsx != 0, x != -1Denominators cannot equal zero, so x cannot be 0 or -1.
2. Combine the fractions on the left-hand side((x + 1) + x) / (x(x + 1)) = 5/6Find a common denominator, which is x(x+1), and add the numerators.
3. Simplify the numerator(2x + 1) / (x^2 + x) = 5/6Combine the like terms in the numerator and expand the denominator.
4. Cross-multiply to eliminate the denominators6(2x + 1) = 5(x^2 + x)Multiply both sides by 6(x^2+x) to clear fractions.
5. Expand both sides12x + 6 = 5x^2 + 5xDistribute 6 on the left and 5 on the right.
6. Rearrange into standard quadratic form5x^2 - 7x - 6 = 0Subtract 12x and 6 from both sides so that one side equals zero.
7. Factor the quadratic equation(5x + 3)(x - 2) = 0Find factors of 5(-6) = -30 that sum to -7, which are -10 and 3, then factor by grouping.
8. Solve for xx = 2 or x = -3/5Set each factor equal to zero: 5x + 3 = 0 gives x = -3/5, and x - 2 = 0 gives x = 2.

Answer: x = 2, x = -3/5

Check: Substitute x = 2: 1/2 + 1/3 = 3/6 + 2/6 = 5/6. Substitute x = -3/5: 1/(-3/5) + 1/((-3/5)+1) = -5/3 + 1/(2/5) = -5/3 + 5/2 = -10/6 + 15/6 = 5/6. Both solutions satisfy the equation and are within the domain.

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4. Solve (x + 1)/(x - 2) = 3/(x - 2) for x

1. Identify restrictions on the variablex - 2 != 0 implies x != 2The denominator cannot equal zero, which means x = 2 is not in the domain.
2. Multiply both sides by the common denominatorx + 1 = 3Multiply both sides by (x - 2) to eliminate the denominators.
3. Solve for xx = 3 - 1 = 2Subtract 1 from both sides to isolate x.
4. Check for extraneous solutionsx = 2 is extraneous because x != 2Since this answer makes the original denominators zero, it must be rejected.

Answer: No solution

Check: Substituting x = 2 into the original equation yields division by zero: (2 + 1)/(2 - 2) = 3/0, which is undefined. Therefore, there are no valid solutions.

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Mistakes the steps help you catch

• Dropping a bracket: in (2x − 1)/3 − (x + 2)/4, multiplying by 12 gives 4(2x − 1) − 3(x + 2), and the minus sign changes the second bracket.
• Cross-multiplying when a side has several terms. It only works when each side is a single fraction.

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