SAT Nonlinear Equations & Systems Question 190 | Free SAT Prep

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Question 190·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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\frac{1}{s}+\frac{1}{s-p}=\frac{3}{p}

The equation above relates real numbers $p$ and $s$ , where $p\ne 0$ , $s\ne 0$ , and $s\ne p$ . What is the sum of all possible values of $s/p$ ?

When an SAT equation involves two variables but the question asks for a ratio like $s/p$ , introduce a single variable for that ratio (for example, $x = s/p$ ) and rewrite the equation entirely in terms of $x$ . Carefully substitute, factor common terms to simplify denominators, and clear fractions to obtain a polynomial—often a quadratic—in $x$ . If you are asked for the sum or product of the possible values, use Vieta’s formulas (sum of roots $= -b/a$ , product $= c/a$ for $ax^2 + bx + c = 0$ ) instead of solving for each root individually; this is faster and reduces algebra mistakes.

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Focus on the ratio, not the individual variables

The question never asks for $s$ or $p$ separately; it asks for $s/p$ . Try defining a new variable to represent this ratio so you can work with just one variable.

Substitute and simplify carefully

After you let $x = s/p$ , write $s$ in terms of $p$ and $x$ , substitute into the original equation, and factor out any common factors in the denominators. What equation do you get that involves only $x$ ?

Remove denominators and look for a quadratic

From your equation in $x$ (it should look like a sum of two fractions equals 3), combine the fractions into a single fraction and then clear the denominator. You should end up with a quadratic equation in $x$ .

Use root-coefficient relationships

Once you have a quadratic equation in the form $ax^2 + bx + c = 0$ , remember that the sum of its roots is $-b/a$ . You do not need to find each root individually to answer this question.

Desmos Guide

Rewrite the equation in terms of the ratio

From the algebra steps, you should get the equation in $x = s/p$ as $1/x + 1/(x-1) = 3$ . You will graph this in Desmos.

Graph both sides of the equation

In Desmos, enter y = 1/x + 1/(x - 1) on one line and y = 3 on another line. Adjust the window if needed so you can see where the two graphs intersect.

Find the possible values of the ratio and their sum

Tap each intersection point of the two graphs and note the x-coordinates; these are the possible values of $s/p$ . Then, in a new expression line, type the sum of those two x-values. The result is the sum of all possible values of $s/p$ .

Step-by-step Explanation

Introduce a single variable for the ratio

We are asked about the ratio $s/p$ , so define a new variable:

Let $x = s/p$ .
Then $s = xp$ .

Substitute $s = xp$ into the original equation

\frac{1}{s} + \frac{1}{s-p} = \frac{3}{p}

to get

\frac{1}{xp} + \frac{1}{xp - p} = \frac{3}{p}.

Factor $p$ out of the second denominator: $xp - p = p(x - 1)$ , so

\frac{1}{xp} + \frac{1}{p(x - 1)} = \frac{3}{p}.

Simplify to an equation in x only

Write each term so the factor $1/p$ is clear:

$\dfrac{1}{xp} = \dfrac{1}{p}\cdot\dfrac{1}{x}$
$\dfrac{1}{p(x-1)} = \dfrac{1}{p}\cdot\dfrac{1}{x-1}$

So the equation becomes

\frac{1}{p}\left(\frac{1}{x} + \frac{1}{x-1}\right) = \frac{3}{p}.

Multiply both sides by $p$ (which is not 0):

\frac{1}{x} + \frac{1}{x-1} = 3.

Now everything is in terms of $x = s/p$ .

Combine the fractions and form a quadratic

Combine the fractions on the left using the common denominator $x(x-1)$ :

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

So we have

\frac{2x - 1}{x(x-1)} = 3.

Cross-multiply to clear the denominator:

2x - 1 = 3x(x-1).

Expand and bring all terms to one side:

\begin{aligned} 2x - 1 &= 3x^2 - 3x \\ 0 &= 3x^2 - 3x - 2x + 1 \\ 0 &= 3x^2 - 5x + 1. \end{aligned}

So $x = s/p$ must satisfy the quadratic equation

3x^2 - 5x + 1 = 0.

Use the quadratic’s coefficients to find the sum of all possible x

The equation $3x^2 - 5x + 1 = 0$ has two roots, and these roots are exactly the possible values of $x = s/p$ .

For a quadratic $ax^2 + bx + c = 0$ , the sum of the roots is $-b/a$ .

Here $a = 3$ and $b = -5$ .
So the sum of the roots (the sum of all possible values of $s/p$ ) is

-\frac{b}{a} = -\frac{-5}{3} = \frac{5}{3}.

Thus, the sum of all possible values of $s/p$ is $\boxed{\frac{5}{3}}$ .

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Step-by-step Explanation

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Step-by-step Explanation