In the xy-plane, the graph of the equation ax + by = 24, where a and b are constants, is a… — SAT

In the $xy$ -plane, the graph of the equation $ax + by = 24$ , where $a$ and $b$ are constants, is a line. The line has an $x$ -intercept of $(d, 0)$ and a $y$ -intercept of $(0, d - 6)$ , where $d > 6$ is a constant. If the line passes through the point $(2, 5)$ , what is the value of $a + b$ ?

Answer: 6

Answer

Substituting the intercepts

(d, 0)

and

(0, d-6)

into

ax + by = 24

gives

a = 24/d

and

b = 24/(d-6)

. Using the point

(2, 5)

, we get

2(24/d) + 5(24/(d-6)) = 24

, which simplifies to

2/d + 5/(d-6) = 1

. Solving for

d

yields

d^2 - 13d + 12 = 0

, giving

d = 12

d = 1

. Given

d > 6

, we have

d = 12

. Substituting

d = 12

back gives

a = 2

and

b = 4

, so

a + b = 6

Step-by-Step Solution

Find expressions for

a

and

b

in terms of

d

by substituting the given intercepts.

a = \frac{24}{d}

and

b = \frac{24}{d-6}

The

x

-intercept

(d, 0)

and

y

-intercept

(0, d-6)

lie on the line

ax + by = 24

Substitute the point

(2, 5)

into the line's equation.

2a + 5b = 24

The point

(2, 5)

lies on the line.

Substitute the expressions for

a

and

b

into

2a + 5b = 24

and simplify.

\frac{2}{d} + \frac{5}{d-6} = 1

To create a single equation in terms of the variable

d

Solve the rational equation for

d

d = 12

Multiplying by the common denominator

d(d-6)

leads to the quadratic equation

d^2 - 13d + 12 = 0

, which factors as

(d-12)(d-1) = 0

. Since the problem states

d > 6

, the only valid solution is

d = 12

Calculate

a

and

b

using

d = 12

, and find their sum.

a = 2

b = 4

, and

a + b = 6

Substituting

d = 12

into the expressions for

a

and

b

yields the constant coefficients, and summing them provides the final requested value.

Key Concept

Using intercepts and a given point on a line in the coordinate plane to solve for the parameters of its linear equation.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic