Let and be distinct real numbers, and let . Solve for (with and ) in terms of , , and if Which of the following gives all possible values of ?

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SAT Nonlinear Equations & Systems Question 162 | Free SAT Prep | Aniko

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Question 162·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Let $a$ and $b$ be distinct real numbers, and let $c\ne 0$ . Solve for $x$ (with $x\ne a$ and $x\ne b$ ) in terms of $a$ , $b$ , and $c$ if

\frac{1}{x-a}+\frac{1}{x-b}=\frac{1}{c}.

Which of the following gives all possible values of $x$ ?

For rational equations like this, first combine the fractions into a single fraction with a common denominator, then clear denominators by cross-multiplying to avoid working with complex fractions. After expanding and simplifying, you almost always get a quadratic; write it in standard form and carefully identify $A$ , $B$ , and $C$ to use the quadratic formula. When simplifying the discriminant, expand fully and look for cancellations and familiar patterns such as $(a\pm b)^2$ , which makes it easier to compare directly with the answer choices.

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Clear the denominators on the left

First, rewrite $\dfrac{1}{x-a} + \dfrac{1}{x-b}$ as a single fraction. What is the common denominator, and what is the numerator after you combine them?

Turn the equation into a quadratic

After you combine the left side into one fraction, set it equal to $\dfrac{1}{c}$ and cross-multiply to eliminate denominators. Expand $(x-a)(x-b)$ and collect like terms to get a quadratic equation in $x$ .

Use the quadratic formula

Once your equation is in the form $x^2 + Bx + C = 0$ , identify $A$ , $B$ , and $C$ , and apply the quadratic formula $x = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ .

Simplify the discriminant carefully

When simplifying $B^2 - 4AC$ , expand the square completely, distribute the $-4$ , and look for terms that cancel. See if the remaining expression can be written using $(a-b)^2$ plus or minus something involving $c^2$ .

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Choose sample values for the parameters

Pick specific real numbers for $a$ , $b$ , and $c$ that satisfy the conditions (for example, $a=1$ , $b=3$ , $c=2$ ). Define these in Desmos by typing a=1, b=3, c=2 (or any other choices with $a\ne b$ and $c\ne 0$ ).

Graph the original equation

Enter the equation 1/(x-a) + 1/(x-b) = 1/c into Desmos. Desmos will show the solution points on the $x$ -axis where this equation holds (you may need to click on the intersection points to see their $x$ -values).

Compute the candidate solutions from an answer choice

For a given answer choice, define its two possible $x$ -values in Desmos using your chosen $a$ , $b$ , and $c$ . For example, for a generic form ((a+b)+K ± sqrt(M))/2, type two expressions: x1 = ((a+b)+K + sqrt(M))/2 and x2 = ((a+b)+K - sqrt(M))/2, replacing K and M according to that choice.

Check which choice matches the true solutions

Compare the numerical values of x1 and x2 from each answer choice with the $x$ -values where the graph of 1/(x-a) + 1/(x-b) = 1/c is true. The correct choice is the one whose two values exactly match the solutions from the graph for different reasonable sets of $(a,b,c)$ .

Step-by-step Explanation

Combine the fractions on the left side

Start with

\frac{1}{x-a} + \frac{1}{x-b} = \frac{1}{c}.

Get a common denominator on the left:

\frac{1}{x-a} + \frac{1}{x-b} = \frac{(x-b) + (x-a)}{(x-a)(x-b)} = \frac{2x - (a+b)}{(x-a)(x-b)}.

So the equation becomes

\frac{2x - (a+b)}{(x-a)(x-b)} = \frac{1}{c}.

Cross-multiply and write a quadratic in standard form

Cross-multiply:

c\big(2x - (a+b)\big) = (x-a)(x-b).

Expand the right-hand side:

(x-a)(x-b) = x^2 - (a+b)x + ab.

So we have

c(2x - a - b) = x^2 - (a+b)x + ab.

Bring everything to one side:

0 = x^2 - (a+b)x + ab - c(2x - a - b).

Distribute the $c$ and combine like terms:

0 = x^2 - (a+b)x + ab - 2cx + c(a+b)

which is

0 = x^2 - \big((a+b)+2c\big)x + \big(ab + c(a+b)\big).

This is a quadratic in $x$ with

$A = 1$ ,
$B = -\big((a+b)+2c\big)$ ,
$C = ab + c(a+b)$ .

Apply the quadratic formula and set up the discriminant

For a quadratic $Ax^2 + Bx + C = 0$ , the solutions are

x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.

Here, $A=1$ , $B = -\big((a+b)+2c\big)$ , and $C = ab + c(a+b)$ , so

-B = (a+b)+2c.

Thus

x = \frac{(a+b)+2c \pm \sqrt{\big((a+b)+2c\big)^2 - 4\big(ab + c(a+b)\big)}}{2}.

Now we just need to simplify the discriminant $D$ :

D = \big((a+b)+2c\big)^2 - 4\big(ab + c(a+b)\big).

Simplify the discriminant and match the answer choice

Simplify $D$ step by step:

\begin{aligned} D &= \big((a+b)+2c\big)^2 - 4\big(ab + c(a+b)\big) \\ &= (a+b)^2 + 4c(a+b) + 4c^2 - 4ab - 4c(a+b) \\ &= (a+b)^2 - 4ab + 4c^2 \\ &= (a^2 + 2ab + b^2 - 4ab) + 4c^2 \\ &= (a^2 - 2ab + b^2) + 4c^2 \\ &= (a-b)^2 + 4c^2. \end{aligned}

So the solutions are

x = \frac{(a+b)+2c \pm \sqrt{(a-b)^2 + 4c^2}}{2}.

This matches answer choice A.

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