Question 62·SAT Desmos Regression Practice·Advanced Math

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For $x \ne 1$ and $x \ne -2$ , the expression

\frac{2x^2+3x-2}{x-1}-\frac{x^2-4}{x+2}+\frac{ax+b}{x^2+x-2}

can be written in the form $x+7$ for all such $x$ , where $a$ and $b$ are constants. Which choice is the value of $a+b$ ?

The phrase “for all such $x$ ” signals an identity: the two expressions agree at every allowed input. A Desmos list regression can use two allowed inputs to find $a$ and $b$ . An excluded input cannot go into the original fractions, but it can become useful after you clear the denominators. Keep that distinction in mind when choosing inputs.

Hints

Hint 1
An identity is an equality that holds at every allowed input. The phrase “can be written in the form” lets you set the entire fraction expression equal to $x+7$ .
Hint 2
For a regression, use $x_1=[0,2]$ . Both inputs avoid zero denominators; at $0$ , the numerator $ax+b$ becomes $b$ . Desmos can use the two inputs to fit both constants.
Hint 3
Desmos will display the fitted values under PARAMETERS. The question asks for $a+b$ , not either constant alone, so enter that sum on a new line.

Step-by-step

Approach 1: Fit the constants in Desmos
Step 1
Turn the equivalent form into an equality
Let $E(x)$ stand for the fraction expression in the question. “For all such $x$ ” means $E(x)=x+7$ is an identity: the equality holds at every input the original fractions allow. So two allowed inputs can give us equations for $a$ and $b$ .
Step 2
Fit both unknown constants
Pick $0$ because it makes $ax+b$ become $b$ , and $2$ because it makes $x^2-4$ zero. Neither is excluded. Type the two Desmos lines shown. Replacing $x$ with $x_1$ lets Desmos use the list of inputs; $\sim$ tells its regression to fit the constants. Under PARAMETERS, Desmos shows $a=-3$ and $b=-6$ .
Step 3
Find the requested sum
Type $a+b$ on the next line. Desmos prints $-9$ . Don't stop at $a=-3$ : the question asks for the sum of both constants, so $a+b=-9$ . Choice B.

Approach 2: Clear denominators and use a strategic input
Step 1
Find the shared denominator
Factor the denominator of the last fraction: $x^2+x-2=(x-1)(x+2)$ . So every denominator divides $(x-1)(x+2)$ , the common denominator you can use to remove all three fractions.
Step 2
Make a polynomial equality
For the allowed inputs, $(x-1)(x+2)$ isn't zero. Multiply both sides of the equality by it, canceling each whole denominator: $(2x^2+3x-2)(x+2)-(x^2-4)(x-1)+(ax+b)=(x+7)(x-1)(x+2)$ . Both sides are now polynomials, with no division left.
Step 3
Use the input that reveals the sum
These polynomials agree at every allowed input, so they're the same polynomial and also agree at $x=1$ . You couldn't put $1$ into the original fractions. Here you can: it makes $ax+b=a+b$ , while every term with $x-1$ vanishes. Substitute: $(2(1)^2+3(1)-2)(1+2)-((1)^2-4)(1-1)+(a(1)+b)=(1+7)(1-1)(1+2)$ . Simplify: $9+a+b=0$ . Subtract $9$ : $a+b=-9$ . The requested sum is $-9$ . Choice B.

Question 62·SAT Desmos Regression Practice·Advanced Math

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