Solve x + 12/x = 7
Rational equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = 3, x = 4 |
Step-by-step solution
12 steps-
1 Givenx + \frac{12}{x} = 7
Solve for x.
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2 Note the values that make a denominator zero — they can never be solutions
Dividing by zero is undefined, so a value that makes a denominator 0 can never be a solution, even if it turns up as an answer later.
x \neq 0 -
3 Multiply every term by the least common denominator x
Multiplying every term on both sides by the least common denominator x cancels every fraction at once. The equation stays balanced because all terms were multiplied by the same thing.
x \left(x + \frac{12}{x}\right) = x 7This clears all the fractions.
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4 Cancel and simplifyx^{2} + 12 = 7 x
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5 Move every term to the left side so the right side is 0
Factoring and the quadratic formula both work on an equation of the form … = 0. Subtracting the right side from both sides gets there without changing the solutions.
x^{2} - 7 x + 12 = 0 -
6 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = -7 and c = 12, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -7,\quad c = 12 -
7 Factor the trinomial: find two numbers whose product is c = 12 and whose sum is b = -7
The goal is to write x² + bx + c as (x + p)(x + q). Multiplying that out gives x² + (p + q)x + p·q, so p and q must multiply to c = 12 and add up to b = -7.
List the pairs of numbers whose product is 12, with their signs, and pick the pair whose sum is -7.
(-4) \cdot (-3) = 12,\qquad (-4) + (-3) = -7 -
8 Write the factored form\left(x - 4\right) \left(x - 3\right) = 0
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9 Zero product property: a product is 0 only when one of its factors is 0
0 is the only number with this property: if a·b = 0, then a = 0 or b = 0. That is why the equation was first rearranged to … = 0 and factored: now each factor can be set to 0 on its own, giving a simpler equation for each.
x - 4 = 0\quad \text{or} \quad x - 3 = 0 -
10 Add 4 to both sides
An equation stays true as long as you do the same thing to both sides, like adding the same weight to both pans of a balance. Adding or subtracting the same amount on both sides moves a term to the other side, where it appears with the opposite sign.
x = 4Move the constant terms to the right.
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11 Add 3 to both sidesx = 3
Move the constant terms to the right.
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12 Check\begin{aligned}x = 3:\quad 7 = 7\quad\checkmark\\ x = 4:\quad 7 = 7\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| 3 | 7 | 7 | ✓ |
| 4 | 7 | 7 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.