Question 54·SAT Desmos Regression Practice·Algebra

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The equation

\frac{1}{2}(k-1)(6x-8) - (k+3) = (2k+5)\left(x - \frac{4}{3}\right) + m,

where $k$ and $m$ are constants, has infinitely many solutions.

Which of the following must be true?

I. $k=8$

II. $m=-11$

III. $m=\dfrac{-7k+23}{3}$

The phrase infinitely many solutions signals a linear identity: both sides must give the same result for every $x$ . Use a Desmos identity regression to find the constants, then compare the factors multiplying $x$ and the terms without $x$ to explain why the values work. Matching only the factors of $x$ isn't enough; the terms without $x$ must match too.

Hints

Hint 1
An identity is true for every $x$ . Two linear expressions that agree at two different inputs must be the same expression. Try a Desmos list of distinct inputs to find the constants that make both sides agree.
Hint 2
A coefficient is the factor multiplying $x$ . After finding $k$ and $m$ , distribute on both sides. Which equation do you get when you match the coefficients of $x$ ?
Hint 3
A constant term has no $x$ . Distributing across $x-\frac{4}{3}$ creates one on the right. Match the whole constant term on each side, then write $m$ in terms of $k$ .

Step-by-step

Fit the identity, then match its terms
Step 1
Find the constants that make both sides identical
Because the sides are linear in $x$ , agreement at two distinct inputs makes them the same expression for every $x$ . Type $x_1=[0,1,2]$ , then enter the given equation with $x_1$ in place of $x$ and $\sim$ in place of $=$ . The $\sim$ tells Desmos to fit the unknown constants. Under PARAMETERS, it shows $k=8$ and $m=-11$ , the values in I and II.
Step 2
Find the left side's constant term
Distribute the products on the left:
$\frac{1}{2}(k-1)(6x-8)-(k+3)=3(k-1)x-4(k-1)-k-3.$
Distribute the remaining $-4$ :
$=3(k-1)x-4k+4-k-3.$
Combine the parts without $x$ :
$=3(k-1)x-5k+1.$
The constant term, the part without $x$ , is $-5k+1$ , not merely $-(k+3)$ .
Step 3
Find the right side's constant term
Distribute $(2k+5)$ on the right:
$(2k+5)\left(x-\frac{4}{3}\right)+m=(2k+5)x-\frac{4}{3}(2k+5)+m.$
So the right constant term includes both $-\frac{4}{3}(2k+5)$ and $m$ .
Step 4
Confirm why k must equal 8
A coefficient is the factor multiplying $x$ . For the equation to hold for every $x$ , those factors must match:
$3(k-1)=2k+5.$
Distribute the $3$ :
$3k-3=2k+5.$
Subtract $2k$ :
$k-3=5.$
Add $3$ :
$k=8.$
Step 5
Match the constant terms to find the rule for m
The equation also needs matching constant terms:
$-5k+1=-\frac{4}{3}(2k+5)+m.$
Add $\frac{4}{3}(2k+5)$ to both sides:
$m=-5k+1+\frac{4}{3}(2k+5).$
Distribute the $\frac{4}{3}$ :
$m=-5k+1+\frac{8k+20}{3}.$
Put all terms over $3$ :
$m=\frac{-15k+3+8k+20}{3}.$
Combine like terms:
$m=\frac{-7k+23}{3}.$
Step 6
Check the rule against the fitted value
Type $\frac{-7k+23}{3}$ beneath the regression. Desmos uses its fitted $k$ and prints $-11$ , the same value it found for $m$ . So III holds along with I and II, making both sides identical for every $x$ . Choice D.

Question 54·SAT Desmos Regression Practice·Algebra

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