Question

assignment: page 348
use the substitution method to determine the solution of each system of linear
equations. check your solutions.
\\(\boldsymbol{\text{a } \

$$\begin{cases} 9x + y = 16 \\\\ y = 7x \\end{cases}$$

}\\) \\(\boldsymbol{\text{b } \

$$\begin{cases} 3x + \\frac{1}{2}y = -3.5 \\\\ y = -6x + 11 \\end{cases}$$

}\\)
\\(\boldsymbol{\text{c } \

$$\begin{cases} y = -5x \\\\ 21x - 7y = 28 \\end{cases}$$

}\\) \\(\boldsymbol{\text{d } \

$$\begin{cases} 2x + 4y = -32 \\\\ y = -\\frac{1}{2}x - 8 \\end{cases}$$

}\\)
\\(\boldsymbol{7}\\) fill in the blank \\(\boldsymbol{1}\\) point
a. the solution is ( type your answer... , type your answer... ).
\\(\boldsymbol{8}\\) multiple choice \\(\boldsymbol{2}\\) points
b. there is ____
\\(\circ\\) only one solution.
\\(\circ\\) no solution.
\\(\circ\\) an infinite number of solutions.

Explanation:

Part (a)

Step1: Substitute \( y = 7x \) into \( 9x + y = 16 \)

Substitute \( y \) in the first equation: \( 9x + 7x = 16 \)

Step2: Solve for \( x \)

Combine like terms: \( 16x = 16 \)
Divide both sides by 16: \( x = \frac{16}{16} = 1 \)

Step3: Solve for \( y \)

Substitute \( x = 1 \) into \( y = 7x \): \( y = 7(1) = 7 \)

Step4: Check the solution

Substitute \( x = 1 \) and \( y = 7 \) into \( 9x + y = 16 \): \( 9(1) + 7 = 9 + 7 = 16 \), which matches the right - hand side.
Substitute into \( y = 7x \): \( 7 = 7(1) \), which is also true.

Step1: Substitute \( y=-6x + 11\) into \( 3x+\frac{1}{2}y=-3.5\)

Substitute \( y \) in the first equation: \( 3x+\frac{1}{2}(-6x + 11)=-3.5 \)

Step2: Simplify the left - hand side

First, distribute \( \frac{1}{2} \): \( 3x-3x+\frac{11}{2}=-3.5 \)
Combine like terms: \( \frac{11}{2}=-3.5 \)
But \( \frac{11}{2}=5.5
eq - 3.5 \), so there is a contradiction.

Step1: Substitute \( y = - 5x\) into \( 21x-7y = 28\)

Substitute \( y \) in the second equation: \( 21x-7(-5x)=28 \)

Step2: Simplify the left - hand side

Distribute - 7: \( 21x + 35x=28 \)
Combine like terms: \( 56x = 28 \)

Step3: Solve for \( x \)

Divide both sides by 56: \( x=\frac{28}{56}=\frac{1}{2} \)

Step4: Solve for \( y \)

Substitute \( x = \frac{1}{2}\) into \( y=-5x \): \( y=-5\times\frac{1}{2}=-\frac{5}{2} \)

Step5: Check the solution

Substitute \( x=\frac{1}{2}\) and \( y = -\frac{5}{2}\) into \( 21x-7y \):
\( 21\times\frac{1}{2}-7\times(-\frac{5}{2})=\frac{21 + 35}{2}=\frac{56}{2}=28 \), which matches the right - hand side.
Substitute into \( y=-5x \): \( -\frac{5}{2}=-5\times\frac{1}{2} \), which is true.

Answer:

The solution is \( (1, 7) \)

Part (b)