In the xy-plane, the graph of the linear equation kx - 3y = 18, where k is a constant, has… — SAT

In the $xy$ -plane, the graph of the linear equation $kx - 3y = 18$ , where $k$ is a constant, has a $y$ -intercept of $(0, b)$ and an $x$ -intercept of $(a, 0)$ , where $a$ and $b$ are constants. If $a - b = 15$ , what is the value of $k$ ?

Answer: 2

Answer

To find the value of

k

, we first determine the intercepts of the linear equation

kx - 3y = 18

. The

y

-intercept

(0, b)

occurs where

x = 0

. Substituting

x = 0

gives

-3b = 18

, which solves to

b = -6

. Next, we use the given relation

a - b = 15

. Substituting

b = -6

yields

a - (-6) = 15

, or

a + 6 = 15

, which gives

a = 9

. The

x

-intercept is therefore

(9, 0)

. Since this point lies on the line, we substitute

x = 9

and

y = 0

into the original equation:

k(9) - 3(0) = 18

. This simplifies to

9k = 18

, which gives

k = 2

Step-by-Step Solution

Find the

y

-coordinate of the

y

-intercept (

b

) by setting

x = 0

in the equation

kx - 3y = 18

b = -6

The

y

-intercept occurs where the graph crosses the

y

-axis, which corresponds to

x = 0

Use the equation

a - b = 15

and the value of

b = -6

to solve for

a

a = 9

Substituting

b = -6

into

a - b = 15

gives

a - (-6) = 15

, which simplifies to

a + 6 = 15

Find the value of

k

by substituting the

x

-intercept

(9, 0)

into the equation

kx - 3y = 18

k = 2

Since

(a, 0) = (9, 0)

is the

x

-intercept, it must satisfy the equation of the line.

Key Concept

Determining intercepts of a linear equation in two variables and using them to find unknown constants.

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