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Chapter 4: Problem 31

(a) solve by substitution. (b) if there is one solution, check. $$ \begin{aligned} &y=-8 x \\ &12 x+5 y=7 \end{aligned} $$

Short Answer

Expert verified
x = -\frac{1}{4}, y = 2. Verification confirms the solution.

Step by step solution

01

Substitute Expression

Start with the first equation: \( y = -8x \). Substitute \( -8x \) for \( y \) in the second equation, which is \( 12x + 5y = 7 \).
02

Substitute and Simplify

After substituting, the second equation becomes: \[ 12x + 5(-8x) = 7 \]. Simplify the equation: \[ 12x - 40x = 7 \].
03

Solve for x

Combine like terms: \[ 12x - 40x = -28x \]. Thus, we have \[ -28x = 7 \]. Divide both sides by -28: \[ x = \frac{7}{-28} = -\frac{1}{4} \].
04

Solve for y

Use the value of \( x \) to find \( y \). Substitute \( -\frac{1}{4} \) into the first equation: \( y = -8(-\frac{1}{4}) \). Therefore, \( y = 2 \).
05

Check the Solution

Substitute \( x = -\frac{1}{4} \) and \( y = 2 \) into the second equation to verify: \[ 12(-\frac{1}{4}) + 5(2) = 7 \]. This simplifies to \[ -3 + 10 = 7 \], which confirms the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Equations
Linear equations are an essential component of algebra and are used to represent straight lines in the coordinate plane. A linear equation in two variables generally takes the form:
\( ax + by = c \).
Here, \( a \), \( b \), and \( c \) are constants. In our exercise, we have two linear equations:
\( y = -8x \)
\( 12x + 5y = 7 \).
Solving these kinds of equations often involves finding the intersection point of the lines they represent, which means finding the values of \( x \) and \( y \) that satisfy both equations simultaneously.
Substitution Method
The substitution method is a helpful technique for solving systems of linear equations. This method involves replacing one variable with an expression involving the other variable. Start by solving one of the equations for one variable. In our exercise, we begin with the equation:
\( y = -8x \).
Next, substitute this expression for \( y \) into the other equation. Doing this allows us to work with a single variable. For our exercise, substituting \( y = -8x \) into \( 12x + 5y = 7 \) yields:
\( 12x + 5(-8x) = 7 \).
This simplifies to \( 12x - 40x = 7 \).
Solving for x and y
After we've substituted and simplified the equations, we can solve for the variable. For our problem, combining like terms resulted in:
\( -28x = 7 \).
To isolate \( x \), divide both sides by -28, giving us:
\( x = \frac{7}{-28} = -\frac{1}{4} \).
With the value of \( x \) found, we can substitute it back into the first equation to find \( y \). Substitute \( x = -\frac{1}{4} \) into \( y = -8x \):
\( y = -8(-\frac{1}{4}) = 2 \).
Hence, \( x = -\frac{1}{4} \) and \( y = 2 \) are our solutions.
Checking Solutions
It's crucial to check our solutions to ensure they satisfy both original equations. Substitute \( x = -\frac{1}{4} \) and \( y = 2 \) back into the second equation to verify:
\( 12(-\frac{1}{4}) + 5(2) = 7 \).
This simplifies to:
\( -3 + 10 = 7 \), which confirms our solution since both sides of the equation are equal. This step of checking the solutions is important as it validates the accuracy of the substitution and calculations. If both equations hold true with the found values of \( x \) and \( y \), we can be confident in our solutions.

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