Question 159·200 Super-Hard SAT Math Questions·Advanced Math

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If $x^2=j+k$ and $y^2=j+m$ , which choice is equivalent to $x^2y^2-j(x^2+y^2)+j^2$ ?

In an equivalent-expression question, the given equations can help you replace repeated parts before comparing choices. Here, replace the squares, then use Desmos to test the expression and the choices at legal values of the letters. One disagreement rules out equivalence; one agreement alone doesn't prove it. If only one choice survives, factoring gives an exact check.

Hints

Hint 1
The givens tell you what $x^2$ and $y^2$ equal, not what $x$ and $y$ equal. Replace each whole square in the expression so it contains only $j$ , $k$ , and $m$ .
Hint 2
Two equivalent expressions give the same result for every allowed set of values. To rule out a choice, you need only one legal test where it differs from the original.
Hint 3
Try $j=1$ , $k=2$ , and $m=3$ . Both $j+k$ and $j+m$ are nonnegative, so real $x$ and $y$ can have those squares. Which choices give the same output as the original?

Step-by-step

Approach 1: Eliminate choices with a Desmos test
Step 1
Replace the given squares
The givens say $x^2=j+k$ and $y^2=j+m$ , so substitute those whole expressions wherever the squares appear:
$x^2y^2-j(x^2+y^2)+j^2=(j+k)(j+m)-j((j+k)+(j+m))+j^2.$
Step 2
Find the original expression's test value
Choose $j=1$ , $k=2$ , and $m=3$ . These are legal values: $x^2=3$ and $y^2=4$ can both hold for real numbers. Type the three assignments and the substituted expression into Desmos. It prints $6$ , so any equivalent choice must also give $6$ for this test.
Step 3
Compare every choice
Keep those lines and type each choice below them. Desmos prints $6$ , $5$ , $11$ , and $5$ , in that order. Only $km$ gives $6$ ; each other choice fails this legal test, so it cannot be equivalent. The equivalent expression is $km$ . Choice A.

Approach 2: Factor to prove the match
Step 1
Distribute the subtraction
No Desmos needed. Factoring proves the match for every allowed value. The $-j$ multiplies both terms inside the parentheses:
$x^2y^2-j(x^2+y^2)+j^2=x^2y^2-jx^2-jy^2+j^2.$
Step 2
Factor the terms in pairs
Factoring pulls a shared factor out of terms. Group the first two terms and the last two; each group leaves $y^2-j$ :
$x^2y^2-jx^2-jy^2+j^2=(x^2y^2-jx^2)+(-jy^2+j^2)$ $=x^2(y^2-j)-j(y^2-j).$
Step 3
Pull out the shared factor
Both terms now contain $y^2-j$ , so factor it out: $x^2(y^2-j)-j(y^2-j)=(x^2-j)(y^2-j)$ . This is the distributive rule run backward.
Step 4
Use both givens
Subtract $j$ from each given equation, so $x^2-j=k$ and $y^2-j=m$ . Replace both factors: $(x^2-j)(y^2-j)=km$ . So the original expression equals $km$ for every allowed value, not only the Desmos test. Choice A.

Question 159·200 Super-Hard SAT Math Questions·Advanced Math

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