In the xy-plane, the line with equation ax + 4y = 36, where a is a constant, has a y-inter… — SAT

In the $xy$ -plane, the line with equation $ax + 4y = 36$ , where $a$ is a constant, has a $y$ -intercept of $(0, p)$ and an $x$ -intercept of $(q, 0)$ , where $p$ and $q$ are positive integers. If $p + q = 15$ , what is the value of $a$ ?

Answer: 6

Answer

To find the value of

a

, we first determine the

y

-intercept of the line by setting

x = 0

in the equation

ax + 4y = 36

. This gives

4y = 36

, so

y = 9

. Thus, the

y

-intercept is

(0, 9)

, which means

p = 9

. Using the given relationship

p + q = 15

, we substitute

p = 9

to find

q = 6

. The

x

-intercept is therefore

(6, 0)

. Substituting these coordinates back into the line's equation gives

a(6) + 4(0) = 36

, which simplifies to

6a = 36

. Solving for

a

yields

a = 6

Step-by-Step Solution

Set

x = 0

in the equation

ax + 4y = 36

to find the

y

-intercept.

4y = 36 \implies y = 9

, so

p = 9

The

y

-intercept of a graph is the point where

x = 0

Substitute

p = 9

into the equation

p + q = 15

to solve for

q

9 + q = 15 \implies q = 6

We are given that the sum of the

y

-coordinate of the

y

-intercept and the

x

-coordinate of the

x

-intercept is

15

Substitute the

x

-intercept

(6, 0)

into the equation

ax + 4y = 36

to solve for

a

a(6) + 4(0) = 36 \implies 6a = 36 \implies a = 6

Since the

x

-intercept is

(q, 0)

and

q = 6

, the point

(6, 0)

must satisfy the equation of the line.

Key Concept

Finding and using intercepts of a linear equation in two variables.

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