SAT Systems of Two Linear Equations Question 42 | Free SAT Prep | Aniko

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Question 42·Hard·Systems of Two Linear Equations in Two Variables

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\begin{cases} (k-4)x + 2y = 1 \\ 3x + (k-6)y = 5 \end{cases}

In the system of equations above, $k$ is a constant. When the system has no solution, what is the sum of all possible values of $k$ ?

Your answer

(Express the answer as an integer)

For SAT questions asking when a 2×2 linear system has no solution, translate this directly to “the lines are parallel but not the same line.” In standard form $Ax+By=C$ , parallel means $A_1/A_2 = B_1/B_2$ , so set these ratios equal and solve for the parameter (here, $k$ ), usually giving a simple linear or quadratic equation. Then quickly check that the same ratio does not hold for the constants (to avoid the infinitely-many-solutions case), and if multiple parameter values work, add or choose them according to what the question asks (like taking the sum). This approach is fast and avoids doing full substitution or elimination for each case.

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Connect “no solution” to the graphs

Think about what the graphs of two linear equations look like when there is no solution. How are their slopes related?

Use the standard form coefficients

Write the equations in the form $Ax + By = C$ . For two lines to be parallel, how must the ratios of the $x$ - and $y$ -coefficients be related?

Set up and solve an equation in k

Set $\dfrac{k-4}{3} = \dfrac{2}{k-6}$ and solve the resulting equation for $k$ . You should get two values—what do you do with both of them?

Exclude the case of infinitely many solutions

After finding the $k$ -values that make the lines parallel, check whether those values also make the entire equations multiples of each other (including the constants). If they did, you would get infinitely many solutions instead of none.

Desmos Guide

Model the parallel condition as a quadratic in Desmos

In Desmos, treat $k$ as the variable $x$ . Enter the equation (x-4)(x-6) - 6 = 0. You can also graph y = (x-4)(x-6) - 6 to see where it crosses the $x$ -axis.

Find the k-values from the graph or solver

Use the $x$ -intercepts of the graph of y = (x-4)(x-6) - 6 (or the solutions to (x-4)(x-6) - 6 = 0) to read off the two $x$ -values. These correspond to the possible $k$ -values that make the lines parallel.

Verify the lines are not identical (optional check)

For each $k$ -value you found, compute (k-4)/3 and compare it with 1/5 in Desmos (for example, by entering each as separate expressions). They should not be equal, confirming the lines are parallel but not the same line.

Compute the required sum

Finally, add the two $k$ -values you read from Desmos in a new expression (e.g., k1 + k2 using the numbers you found). The result is the number you should enter as your final answer.

Step-by-step Explanation

Translate “no solution” for a linear system

A system of two linear equations has no solution when the two lines are parallel but distinct.

For equations in the form

a_1x + b_1y = c_1 \\[0.7em] a_2x + b_2y = c_2,

parallel lines mean the coefficient ratios match:

\frac{a_1}{a_2} = \frac{b_1}{b_2},

but to avoid being the same line (which would give infinitely many solutions), this common ratio must not equal $c_1/c_2$ .

Set up the condition for parallel lines

From the system

(k-4)x + 2y = 1 \\[0.7em] 3x + (k-6)y = 5,

we identify

$a_1 = k-4$ , $b_1 = 2$ , $c_1 = 1$ ,
$a_2 = 3$ , $b_2 = k-6$ , $c_2 = 5$ .

For the lines to be parallel:

\frac{k-4}{3} = \frac{2}{k-6}.

Now solve this equation for $k$ .

Solve the proportionality for k

Solve

\frac{k-4}{3} = \frac{2}{k-6}.

Cross-multiply:

(k-4)(k-6) = 6.

Expand the left side:

(k-4)(k-6) = k^2 -10k +24.

So we get

k^2 -10k +24 = 6,\\[0.7em] k^2 -10k +18 = 0.

Now solve this quadratic equation for $k$ (by factoring or the quadratic formula).

Find the two k-values and check they do not give the same line

Use the quadratic formula on

k^2 -10k +18 = 0.

Here $a=1$ , $b=-10$ , $c=18$ .

k = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(18)}}{2(1)} = \frac{10 \pm \sqrt{100 - 72}}{2} = \frac{10 \pm \sqrt{28}}{2} = 5 \pm \sqrt{7}.

So there are two $k$ -values: $k = 5+\sqrt{7}$ and $k = 5-\sqrt{7}$ .

To make sure the lines are not the same line, we need

\frac{k-4}{3} \neq \frac{1}{5}.

If you plug in either $k = 5+\sqrt{7}$ or $k = 5-\sqrt{7}$ , the value of $(k-4)/3$ is not $1/5$ , so in both cases the lines are parallel and distinct, giving no solution.

Add the valid k-values

The problem asks for the sum of all possible values of $k$ that make the system have no solution.

We found two such values: $5+\sqrt{7}$ and $5-\sqrt{7}$ . Their sum is

(5+\sqrt{7}) + (5-\sqrt{7}) = 10.

Answer: $10$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation