Question 551·Hard·Algebra

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The equation $(3m - 9)(x + 2) = 6x + 12 - 4m$ is shown above, where $m$ is a real number. For which value of $m$ does the equation have no solution for $x$ ?

A linear equation can have no solution when its $x$ -terms cancel but the numbers left behind disagree. Distribute first, then match the coefficients, the numbers multiplying $x$ , to find a possible value of the unknown constant. Check what remains: matching coefficients alone could instead make the equation true for every $x$ .

Hints

Hint 1
A coefficient is the number multiplying $x$ . Distribute $3m-9$ across $x+2$ to see the left side's coefficient. What must it equal for the $x$ -terms to cancel?
Hint 2
Matching coefficients is only the first test. After the $x$ -terms cancel, compare the constants, the terms without $x$ . Unequal constants mean no value of $x$ can make the equation true.

Step-by-step

Match coefficients, then check
Step 1
Expose the $x$ -coefficients
Distribute the whole factor $3m-9$ to both $x$ and $2$ :
$(3m-9)x+6m-18=6x+12-4m.$
Step 2
Make the $x$ -terms cancel
The coefficients, or numbers multiplying $x$ , are $3m-9$ and $6$ . For no solution, they must match. Set them equal: $3m-9=6$ .Matching coefficients makes the $x$ -terms cancel; the constants must still disagree.
Step 3
Isolate the multiple of $m$
Add $9$ to both sides to leave only the $m$ -term:
$3m=15.$
Step 4
Check the original equation in Desmos
Divide both sides by $3$ : $m=5$ .Type each side of the original equation with $m=5$ as its own graph: $y=(3(5)-9)(x+2)$ and $y=6x+12-4(5)$ . Desmos shows parallel lines crossing the $y$ -axis at $12$ and $-8$ . Their different intercepts mean they never meet, so no $x$ works. The required value of $m$ is $5$ . Choice C.

Question 551·Hard·Algebra

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

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Step-by-step

Lessons that teach this