Question 554·Hard·Algebra

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(3k - 6)x + 4y = 8

9x + ky = 6

The system of equations above contains the real parameter $k$ . What is the sum of all real values of $k$ for which the system has no solution?

Your answer

For a linear system with a parameter, no solution means the two lines are parallel but distinct. Match their variable coefficients to find candidate parameter values, then graph an equation in the parameter to see how many real candidates there are. Check the constants too: proportional coefficients can describe the same line, which has infinitely many solutions. Symmetry can give the candidates’ exact sum.

Hints

Hint 1
For two lines to be parallel, their $x$ - and $y$ -coefficients must share one multiplier. You can avoid dividing by a coefficient that might be $0$ : for coefficient pairs $(A,B)$ and $(C,D)$ , set $AD=BC$ .
Hint 2
Graph the resulting equation with the graph’s $x$ standing for $k$ . Two intersections with a horizontal line give two real candidates. Before counting them, how can you rule out the possibility that the original lines overlap?
Hint 3
A parabola’s axis of symmetry is the vertical line through its middle. If both candidates work, their graph points sit equally far from that axis. What does that tell you about the candidates’ sum?

Step-by-step

Match coefficients, then use graph symmetry
Step 1
Find the condition for parallel lines
For no solution, the two equations must draw distinct parallel lines. Their coefficients, the numbers multiplying $x$ and $y$ , must share one multiplier. Cross-multiply the coefficient pairs to avoid dividing by a value that could be $0$ : $(3k-6)k=4(9)=36$ . This finds the parallel candidates, but it could also include overlapping lines.
Step 2
Make the candidate equation easier to graph
Since $3k-6=3(k-2)$ , divide the candidate equation by $3$ : $k(k-2)=12$ . Every value of $k$ that could make the lines parallel must satisfy this equation.
Step 3
Count the real candidates in Desmos
Type $y=x(x-2)$ and $y=12$ , using the graph’s $x$ -coordinate to stand for $k$ . Click their two intersections: Desmos shows approximately $(-2.606,12)$ and $(4.606,12)$ . So there are two real candidates. The displayed coordinates are rounded, so don’t add them for an exact answer.
Step 4
Find the only way the lines could overlap
If the lines overlap, the constants require the second equation to be $\frac{6}{8}=\frac34$ times the first. Apply that multiplier to the first equation’s $y$ coefficient: $k=\frac34(4)=3$ . So $3$ is the only possible identical-line value.
Step 5
Rule out overlapping lines
Test $k=3$ in the parallel-candidate equation: $3(3-2)=3\ne12$ . It fails the parallel condition, so neither graphed candidate makes the original lines overlap. Both candidates give distinct parallel lines and count as no-solution values.
Step 6
Use symmetry to get the exact sum
The graph of $y=x(x-2)$ has an axis of symmetry at $x=1$ , since $x(x-2)=(x-1)^2-1$ . Its two intersections with $y=12$ sit equally far from $1$ . So both valid $k$ values add to $2(1)=2$ . The requested sum is $2$ . Grid in 2.

Question 554·Hard·Algebra

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this