\begin{aligned} k x - 5y &= 8 \\ 3x - 2y &= 12 \end{aligned} In the system of equations ab… — SAT

\begin{aligned} k x - 5y &= 8 \\ 3x - 2y &= 12 \end{aligned}

In the system of equations above, $k$ is a constant. If the system has no solution, what is the value of $k$ ?

\frac{15}{2}

The correct answer is the value that makes the two lines parallel. Rewriting both equations in slope-intercept form gives the slopes

\frac{k}{5}

and

\frac{3}{2}

. Equating these slopes yields

\frac{k}{5} = \frac{3}{2}

, which simplifies to

\frac{15}{2}

. Since their

y

-intercepts (

-\frac{8}{5}

and

-6

) are distinct, the lines are parallel and never intersect, meaning the system has no solution.

Convert the first equation to slope-intercept form (

y = mx + b

kx - 5y = 8 \implies -5y = -kx + 8 \implies y = \frac{k}{5}x - \frac{8}{5}

This identifies the slope (

m_1 = \frac{k}{5}

) and the

y

-intercept (

b_1 = -\frac{8}{5}

) of the first line.

Convert the second equation to slope-intercept form.

3x - 2y = 12 \implies -2y = -3x + 12 \implies y = \frac{3}{2}x - 6

This identifies the slope (

m_2 = \frac{3}{2}

) and the

y

-intercept (

b_2 = -6

) of the second line.

Equate the two slopes to find the value of

k

that makes the lines parallel.

\frac{k}{5} = \frac{3}{2} \implies k = 5 \times \frac{3}{2} = \frac{15}{2}

A system of linear equations has no solution if the lines are parallel (equal slopes) and distinct (different

y

-intercepts). Since the

y

-intercepts

-\frac{8}{5}

and

-6

are different, setting the slopes equal guarantees no solution.

A system of two linear equations has no solution if the lines represented by the equations have the same slope but different

y

-intercepts (parallel lines).

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