Question 521·Hard·Algebra

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The equation $6(k-2)(x+3)=3(4x+12)$ , where $k$ is a constant, is to be solved for $x$ . For which value of $k$ will the equation have infinitely many real solutions?

An equation with infinitely many solutions must be an identity: both sides must give the same value for every $x$ . Look for a shared factor before expanding, then match the numbers multiplying it and solve for the constant with a Desmos regression. Finding one $x$ that works is not enough.

Hints

Hint 1
A factor is a part of an expression being multiplied. Factor $4$ out of $4x+12$ . What expression inside the parentheses now matches the left side?
Hint 2
For every $x$ to work, matching at one value of $x$ isn't enough. Once both sides contain $x+3$ , what must be true of the numbers multiplying it?

Step-by-step

Match the shared factor
Step 1
Find the factor on the right
Both terms in $4x+12$ are divisible by $4$ . Factor it out:
$4x+12=4(x+3).$
The same factor, $x+3$ , now appears on both sides.
Step 2
Rewrite the right side
Multiply the outside $3$ by $4$ :
$3[4(x+3)]=12(x+3).$
So the right-hand multiplier of $x+3$ is $12$ .
Step 3
Make the equation true for every x
An identity is an equation true for every value of $x$ . Here, the multipliers of $x+3$ must match, since $x+3$ isn't always zero:
$6(k-2)=12.$
At $x=-3$ , both sides are zero for any $k$ . That single solution doesn't mean every $x$ works.
Step 4
Find the constant
Type $6(k-2)\sim12$ in Desmos. The $\sim$ tells Desmos to run a regression; it reports $k=4$ under PARAMETERS. With that value, both sides are $12(x+3)$ for every real $x$ . The required value is $4$ . Choice C.

Question 521·Hard·Algebra

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