Question

problems 5–7: here is a graph that represents one equation in a system of equations.

1. write a second equation for the system so that it has infinitely many solutions.
2. write a second equation whose graph goes through (0, 1) so that the system has no solution.
3. write a second equation whose graph goes through (0, 2) so that the system has one solution: (4, 1).

spiral review

1. select all the equations that have no solution.

□ a. $2 + 4(4x + 5) = 8x + 2x - 11$
□ b. $-x + 3x - 7 = 2(x - 7)$
□ c. $7 - 5x(-3) = 5(3x - 2)$
□ d. $6x + 3(2x - 1) = 5x - 4 + 7x + 1$
problems 9–10: solve each equation. show or explain your thinking.

1. $\frac{15(x - 3)}{5} = 3(2x - 3)$
2. $0.4(x + 7) = 0.2(x + 40) - 5.2 + 0.2x$

Explanation:

Problems 5-7

Step1: Rewrite given equation in slope-intercept

First, rewrite $4y=-3x+16$ to $y=mx+b$ form:
$$y = -\frac{3}{4}x + 4$$

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Problem 5 Step1: Scale original equation

A system has infinitely many solutions if equations are scalar multiples. Multiply original equation by 2:
$$2(4y) = 2(-3x+16) \implies 8y = -6x + 32$$

Problem 6 Step1: Set same slope, new y-intercept

No solution means parallel lines (same slope, different y-intercept). Use $m=-\frac{3}{4}$, $b=1$:
$$y = -\frac{3}{4}x + 1$$
Rearrange to standard form: $3x + 4y = 4$

Problem 7 Step1: Find slope of new line

New line passes through $(0,2)$ and $(4,1)$. Calculate slope:
$$m = \frac{1-2}{4-0} = -\frac{1}{4}$$

Problem 7 Step2: Write line equation

Use $b=2$ (y-intercept) in $y=mx+b$:
$$y = -\frac{1}{4}x + 2$$
Rearrange to standard form: $x + 4y = 8$

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Problem 8

Step1: Simplify Option A

Expand and combine like terms:
$$2 + 16x + 20 = 10x - 11 \implies 16x + 22 = 10x - 11 \implies 6x = -33$$
This has one solution.

Step2: Simplify Option B

Expand and combine like terms:
$$2x - 7 = 2x - 14 \implies -7 = -14$$
This is a false statement (no solution).

Step3: Simplify Option C

Expand and combine like terms:
$$7 + 15x = 15x - 10 \implies 7 = -10$$
This is a false statement (no solution).

Step4: Simplify Option D

Expand and combine like terms:
$$6x + 6x - 3 = 12x - 3 \implies 12x - 3 = 12x - 3$$
This is always true (infinitely many solutions).

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Problem 9

Step1: Eliminate denominator

Multiply both sides by 5:
$$15(x-3) = 15(2x-3)$$

Step2: Divide both sides by 15

Simplify the equation:
$$x - 3 = 2x - 3$$

Step3: Solve for $x$

Isolate $x$ terms:
$$-3 + 3 = 2x - x \implies x=0$$

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Problem 10

Answer:

s:

1. Problem 5: $\boldsymbol{8y = -6x + 32}$ (or any non-zero scalar multiple of $4y=-3x+16$)
2. Problem 6: $\boldsymbol{3x + 4y = 4}$ (or $y=-\frac{3}{4}x+1$)
3. Problem 7: $\boldsymbol{x + 4y = 8}$ (or $y=-\frac{1}{4}x+2$)
4. Problem 8: $\boldsymbol{B. -x + 3x - 7 = 2(x - 7)}$, $\boldsymbol{C. 7 - 5x(-3) = 5(3x - 2)}$
5. Problem 9: $\boldsymbol{x=0}$
6. Problem 10: $\boldsymbol{\text{Infinitely many solutions}}$ (all real numbers)