Q&A for How to Solve Rational Equations

An extraneous solution refers to a result that may appear valid when substituting a certain value for X in an equation, but it often leads to a situation where the expression becomes indeterminate, such as division by zero. When working with the original (parent) function, extraneous solutions may not be immediately apparent, as simplification steps can eliminate them. However, upon revisiting the original function, plugging in an extraneous solution may render the function undefined or result in a zero-over-zero scenario. In essence, extraneous solutions are deemed irrelevant as they don't contribute valid answers to the problem, given that zero over zero or undefined values are not considered part of the solution set.

Solve for x by multiplying both sides of the equation by 4x to get rid of the fractions: 4x(5/4x + 1/x) = 4x(3). Then 5 + 4 = 12x, 9 = 12x, and x = 9/12 = ¾.

If the equation is 6x = 5 - ¾, all you have to do to solve for x is to divide both sides of the equation by 6: 6x = 4¼ = 17/4. Divide by 6: x = 17/24.

From both sides of the equation subtract [(x+5)/(x-5)]. That leaves [(x+3)/(x-3)] = 0. Multiply both sides of the equation by (x-3). That gives you (x+3) = 0. Therefore, x = -3.

Multiply both sides of the equation by 13.

Factor 3 out of both numerator and denominator. The 3's cancel. That leaves a / (1 - 9a).

Multiply both sides of the equation by 15 to "clear" (get rid of) the fractions (just because fractions are not as easy to work with). Now you have 3x - 5x = 30. So -2x = 30. Therefore, x = -15.

A rational number is any number that can be accurately expressed as a simple fraction (that is, one whole number divided by another). Examples are 1 (which is 1/1), 34 (which is 34/1), 2/3, and 48/37.

First, let's get what we can out of the way. 1/3(2m-12) expands to 1/(6m-36). Now let's move around the equation. Multiply both sides by (6m-36) to get rid of the fraction. 2m(6m-36)/5 = 1. That expands to (12m^2 - 72m)/5 = 1. Now let's get rid of the other fraction by multiplying by 5 on both sides. 12m^2 - 72m = 5. Subtract 5 from both sides to become 12m^2 - 72m - 5 = 0. You can then use the quadratic equation.

Cross-multiply (numerator #1, multiplied by denominator #2 = denominator #1, multiplied by numerator #2): x² = 64. So x = +/- 8.

Multiply both sides of the equation by 4x: 24 + 4x² + 3x = 8x. Subtract 8x from both sides: 4x² - 5x + 24 = 0. Because the left side won't factor, use the quadratic formula to solve for x.

x + 1 = 4 / (x + 1). Multiply both sides by (x + 1). That gives you (x + 1)² = 4, or x² + 2x + 1 = 4. Subtract 4 from both sides: x² + 2x - 3 = 0. Factor the left side: (x + 3)(x - 1) = 0. So x = 1 or -3.

No. See the opening definition in the introduction of the article above.

4r / (5r + 6) = 1/6. Multiply both sides by the product of the two denominators: (4r)(6)(5r + 6) / (5r + 6) = (1)(6)(5r + 6) / 6. After canceling, 24r = 5r + 6. Subtract 5r from both sides: 19r = 6. So r = 6/19. (If you substitute 6/19 for r in the original equation, the equation is valid.)

You can't solve it (because it's not an equation), but you can simplify it by factoring both the numerator and the denominator and then canceling common factors: (A³ + 8) / (A² - 4) = (A + 2)(A² - 2A + 4) / (A + 2)(A - 2) = (A² - 2A + 4) / (A - 2). This may not seem any simpler, but it's now a second-degree expression rather than a third-degree (cubic) one.

Let's assume the equation is 1/x = 2 / (3x-1). Cross-multiply: (1)(3x-1) = (2)(x), or 3x-1 = 2x. Then x-1 = 0, and x = 1.

Multiply both sides of the equation by 45 (which is 9 x 5): 5y - 18 = 15. Then 5y = 33, and y = 33/5.

Let's assume the equation is this: 5 / (3x+5) = [3 / [(3x+5)] + 1. Multiply both sides of the equation by (3x+5): 5 = 3 + (3x+5) = 3x + 8. Subtract 8 from both sides: -3 = 3x, and x = -1.

Let's assume the equation is this: (x+2) / 3 = (2x-4) / 2. The right side of the equation simplifies to x-2, so (x+2) / 3 = x-2. Multiply both sides by 3: x+2 = 3x - 6. Subtract x from both sides, and add 6 to both sides: 2+6 = 2x, so 8 = 2x, and x = 4.

Let's assume the equation is: 2 / (x+3) = 3 / (x+4). Multiply both sides of the equation by [(x+3)(x+4)]: 2(x+4) = 3(x+3), so 2x+8 = 3x+9. Subtract 2x from both sides, and subtract 9 from both sides. Then -1 = x.

Let's assume the equation is 4n + 1 + n² - 5n - 6 = n - 6. Combine terms: -n - 5 + n² = n - 6. Subtract n from both sides, and add 6 to both sides: -2n + 1 + n² = 0. Re-order the left side: n² - 2n + 1 = 0. Factor the left side: (n - 1)² = 0. Find the square root of each side: +/-(n - 1) = √0 = 0. In the first case, +(n - 1) = 0, and n = 1. In the second case, -(n - 1) = -n + 1 = 0, and n = 1. So in both cases n = 1.

Multiply both sides by 8: x + 2 = 6. Then x - 4 = 0, and x = 4.

Multiply both sides by the product of the denominators, xy. This simplifies the equation by eliminating the fractions (but doesn't help solve the equation).

It's not an equation, so you can't solve it. It also cannot be simplified, unless you first use parentheses to group terms.

First, multiply both sides of the equation by 4 to get rid of the fraction. Then you have an easily solved equation in x: 8x + 12 = x. Subtract x from both sides and subtract 12 from both sides: 7x = -12, and x = -12/7.

First, multiply both sides of the equation by 15 (which is the product of the denominators 5 and 3). (It's a coincidence that there is a 15 in the equation.) That gets rid of the fractions, and you now have an easily solved equation in x.

That expression cannot be "solved" because it's not an equation.

x/3 = 7x / (12x - 15) = 7x / 3(4x - 5). Multiply both sides by 3: x = 7x / (4x - 5). Multiply both sides by (4x - 5): 7x = x(4x - 5) = 4x² - 5x. Subtract 7x from both sides: 0 = 4x² - 12x = x(4x - 12). Then x = 0 or 3. (Two solutions because the equation is quadratic.)

x/5 = 15 + (x/3). Multiply both sides by (5)(3), or 15. That gives us 3x = 225 + 5x. Subtract 5x from both sides: -2x = 225. Divide both sides by (-2): x = -112.5.

(2/11)(-14/5) = -28/55.