graph each inequality.
46. ( 0 geq -6 - 2x - 5y ) 47. ( y |x| ) 48. ( y geq |x - 3| )
49. a linear inequality has the points ( (0, 3) ) and ( (-3, 1.5) ) as solutions on the boundary line. also, the point ( (1, 1) ) is not a solution. write the linear inequality.
50. two linear inequalities are graphed on the same coordinate plane. the point ( (0, 0) ) is a solution of both inequalities. the entire coordinate plane is shaded except for quadrant i. what are the two inequalities?
spiral review
tell whether each function is linear. if so, graph the function. (lesson 5 - 1)
51. ( y = 2x - 4 ) 52. ( y = x^2 + 2 ) 53. ( y = 3 )
write an equation in slope - intercept form for the line through the two points. (lesson 5 - 7)
54. ( (0, 9) ) and ( (5, 2) ) 55. ( (-5, -2) ) and ( (7, 7) ) 56. ( (0, 0) ) and ( (-8, -10) )
57. ( (-1, -2) ) and ( (1, 4) ) 58. ( (2, 2) ) and ( (6, 5) ) 59. ( (-3, 2) ) and ( (3, -1) )
solve each system by elimination. (lesson 6 - 3)
60. ( \begin{cases}x + 6y = 14\x - 6y = -10end{cases} ) 61. ( \begin{cases}x + y = 13\3x + y = 9end{cases} ) 62. ( \begin{cases}2x - 4y = 18\5x - y = 36end{cases} )
63. ( \begin{cases}2y + x = 12\y - 2x = 1end{cases} ) 64. ( \begin{cases}2y - 6x = -8\y = -5x + 12end{cases} ) 65. ( \begin{cases}2x + 3y = 33\y = \frac{1}{4}xend{cases} )

Question

graph each inequality.
46. ( 0 geq -6 - 2x - 5y ) 47. ( y > |x| ) 48. ( y geq |x - 3| )
49. a linear inequality has the points ( (0, 3) ) and ( (-3, 1.5) ) as solutions on the boundary line. also, the point ( (1, 1) ) is not a solution. write the linear inequality.
50. two linear inequalities are graphed on the same coordinate plane. the point ( (0, 0) ) is a solution of both inequalities. the entire coordinate plane is shaded except for quadrant i. what are the two inequalities?
spiral review
tell whether each function is linear. if so, graph the function. (lesson 5 - 1)
51. ( y = 2x - 4 ) 52. ( y = x^2 + 2 ) 53. ( y = 3 )
write an equation in slope - intercept form for the line through the two points. (lesson 5 - 7)
54. ( (0, 9) ) and ( (5, 2) ) 55. ( (-5, -2) ) and ( (7, 7) ) 56. ( (0, 0) ) and ( (-8, -10) )
57. ( (-1, -2) ) and ( (1, 4) ) 58. ( (2, 2) ) and ( (6, 5) ) 59. ( (-3, 2) ) and ( (3, -1) )
solve each system by elimination. (lesson 6 - 3)
60. ( \begin{cases}x + 6y = 14\x - 6y = -10end{cases} ) 61. ( \begin{cases}x + y = 13\3x + y = 9end{cases} ) 62. ( \begin{cases}2x - 4y = 18\5x - y = 36end{cases} )
63. ( \begin{cases}2y + x = 12\y - 2x = 1end{cases} ) 64. ( \begin{cases}2y - 6x = -8\y = -5x + 12end{cases} ) 65. ( \begin{cases}2x + 3y = 33\y = \frac{1}{4}xend{cases} )

Accepted Answer

### For Questions 46-50 (Inequalities)
# Explanation:
---
#### Question 46: $0 \geq -6 - 2x - 5y$
## Step1: Rearrange to slope-intercept form
$0 +6 +2x \geq -5y \implies 2x +6 \geq -5y \implies y \geq -\frac{2}{5}x - \frac{6}{5}$
## Step2: Graph the boundary line
Boundary line: $y = -\frac{2}{5}x - \frac{6}{5}$ (solid, since $\geq$). Shade above the line.

---
#### Question 47: $y > |x|$
## Step1: Identify the boundary graph
Boundary graph: $y=|x|$ (V-shaped, vertex at (0,0), dashed since $>$)
## Step2: Determine shaded region
Shade all areas **above** the V-shaped graph.

---
#### Question 48: $y \geq |x-3|$
## Step1: Identify the boundary graph
Boundary graph: $y=|x-3|$ (V-shaped, vertex at (3,0), solid since $\geq$)
## Step2: Determine shaded region
Shade all areas **above** the V-shaped graph.

---
#### Question 49
## Step1: Calculate slope of boundary line
Slope $m = \frac{1.5-3}{-3-0} = \frac{-1.5}{-3} = 0.5 = \frac{1}{2}$
## Step2: Write boundary line equation
Using $(0,3)$: $y = \frac{1}{2}x + 3$
## Step3: Test point to find inequality sign
Substitute $(1,1)$: $1 \stackrel{?}{\leq} \frac{1}{2}(1)+3 = 3.5$ (true, but (1,1) is not a solution, so reverse the sign)
Inequality: $y \geq \frac{1}{2}x + 3$

---
#### Question 50
## Step1: Identify Quadrant I constraints
Quadrant I is where $x>0$ and $y>0$. To exclude it, the inequalities must shade everywhere else, with (0,0) as a solution.
## Step2: Determine valid inequalities
$x \leq 0$ (shades left of y-axis, includes (0,0)) and $y \leq 0$ (shades below x-axis, includes (0,0)). Together they shade all except Quadrant I.

# Answer:
46. Boundary line: $y = -\frac{2}{5}x - \frac{6}{5}$ (solid), shade above the line
47. Boundary graph: $y=|x|$ (dashed), shade above the graph
48. Boundary graph: $y=|x-3|$ (solid), shade above the graph
49. $\boldsymbol{y \geq \frac{1}{2}x + 3}$
50. $\boldsymbol{x \leq 0}$ and $\boldsymbol{y \leq 0}$

---

### For Questions 51-53 (Linear Functions)
# Explanation:
---
#### Question 51: $y=2x-4$
## Step1: Check linearity
Follows $y=mx+b$ (degree 1), so linear.
## Step2: Graph key points
Y-intercept: $(0,-4)$; X-intercept: $0=2x-4 \implies x=2$, so $(2,0)$. Draw solid line through points.

---
#### Question 52: $y=x^2+2$
## Step1: Check linearity
Has $x^2$ term (degree 2), so not linear.

---
#### Question 53: $y=3$
## Step1: Check linearity
Follows $y=mx+b$ ($m=0$, $b=3$), so linear.
## Step2: Graph the line
Horizontal solid line at $y=3$, parallel to x-axis.

# Answer:
51. Linear; graph: solid line through $(0,-4)$ and $(2,0)$
52. Not linear
53. Linear; graph: horizontal solid line at $y=3$

---

### For Questions 54-59 (Slope-Intercept Equations)
# Explanation:
---
#### Question 54: $(0,9)$ and $(5,2)$
## Step1: Calculate slope
$m = \frac{2-9}{5-0} = -\frac{7}{5}$
## Step2: Write slope-intercept form
Y-intercept is $(0,9)$, so $y = -\frac{7}{5}x + 9$

---
#### Question 55: $(-5,-2)$ and $(7,7)$
## Step1: Calculate slope
$m = \frac{7-(-2)}{7-(-5)} = \frac{9}{12} = \frac{3}{4}$
## Step2: Solve for b
Substitute $(-5,-2)$: $-2 = \frac{3}{4}(-5)+b \implies -2 = -\frac{15}{4}+b \implies b = -2 + \frac{15}{4} = \frac{7}{4}$
Equation: $y = \frac{3}{4}x + \frac{7}{4}$

---
#### Question 56: $(0,0)$ and $(-8,-10)$
## Step1: Calculate slope
$m = \frac{-10-0}{-8-0} = \frac{10}{8} = \frac{5}{4}$
## Step2: Write slope-intercept form
Y-intercept is $(0,0)$, so $y = \frac{5}{4}x$

---
#### Question 57: $(-1,-2)$ and $(1,4)$
## Step1: Calculate slope
$m = \frac{4-(-2)}{1-(-1)} = \frac{6}{2} = 3$
## Step2: Solve for b
Substitute $(1,4)$: $4 = 3(1)+b \implies b=1$
Equation: $y=3x+1$

---
#### Question 58: $(2,2)$ and $(6,5)$
## Step1: Calculate slope
$m = \frac{5-2}{6-2} = \frac{3}{4}$
## Step2: Solve for b
Substitute $(2,2)$: $2 = \frac{3}{4}(2)+b \implies 2 = \frac{3}{2}+b \implies b = \frac{1}{2}$
Equation: $y = \frac{3}{4}x + \frac{1}{2}$

---
#### Question 59: $(-3,2)$ and $(3,-1)$
## Step1: Calculate slope
$m = \frac{-1-2}{3-(-3)} = \frac{-3}{6} = -\frac{1}{2}$
## Step2: Solve for b
Substitute $(3,-1)$: $-1 = -\frac{1}{2}(3)+b \implies -1 = -\frac{3}{2}+b \implies b = \frac{1}{2}$
Equation: $y = -\frac{1}{2}x + \frac{1}{2}$

# Answer:
54. $\boldsymbol{y = -\frac{7}{5}x + 9}$
55. $\boldsymbol{y = \frac{3}{4}x + \frac{7}{4}}$
56. $\boldsymbol{y = \frac{5}{4}x}$
57. $\boldsymbol{y=3x+1}$
58. $\boldsymbol{y = \frac{3}{4}x + \frac{1}{2}}$
59. $\boldsymbol{y = -\frac{1}{2}x + \frac{1}{2}}$

---

### For Questions 60-65 (Systems by Elimination)
# Explanation:
---
#### Question 60: $\begin{cases} x+6y=14 \ x-6y=-10 \end{cases}$
## Step1: Add equations to eliminate y
$(x+6y)+(x-6y)=14+(-10) \implies 2x=4$
## Step2: Solve for x
$x = \frac{4}{2} = 2$
## Step3: Substitute x to find y
$2+6y=14 \implies 6y=12 \implies y=2$

---
#### Question 61: $\begin{cases} x+y=13 \ 3x+y=9 \end{cases}$
## Step1: Subtract first equation from second
$(3x+y)-(x+y)=9-13 \implies 2x=-4$
## Step2: Solve for x
$x = \frac{-4}{2} = -2$
## Step3: Substitute x to find y
$-2+y=13 \implies y=15$

---
#### Question 62: $\begin{cases} 2x-4y=18 \ 5x-y=36 \end{cases}$
## Step1: Multiply second equation by 4
$4(5x-y)=4(36) \implies 20x-4y=144$
## Step2: Subtract first equation
$(20x-4y)-(2x-4y)=144-18 \implies 18x=126$
## Step3: Solve for x
$x = \frac{126}{18} = 7$
## Step4: Substitute x to find y
$5(7)-y=36 \implies 35-y=36 \implies y=-1$

---
#### Question 63: $\begin{cases} x+2y=12 \ -2x+y=1 \end{cases}$
## Step1: Multiply second equation by 2
$2(-2x+y)=2(1) \implies -4x+2y=2$
## Step2: Subtract from first equation
$(x+2y)-(-4x+2y)=12-2 \implies 5x=10$
## Step3: Solve for x
$x = \frac{10}{5} = 2$
## Step4: Substitute x to find y
$2+2y=12 \implies 2y=10 \implies y=5$

---
#### Question 64: $\begin{cases} -6x+2y=-8 \ y=-5x+12 \end{cases}$
## Step1: Substitute y into first equation
$-6x+2(-5x+12)=-8$
## Step2: Simplify and solve for x
$-6x-10x+24=-8 \implies -16x=-32 \implies x=2$
## Step3: Substitute x to find y
$y=-5(2)+12=2$

---
#### Question 65: $\begin{cases} 2x+3y=33 \ y=\frac{1}{4}x \end{cases}$
## Step1: Substitute y into first equation
$2x+3(\frac{1}{4}x)=33$
## Step2: Simplify and solve for x
$2x+\frac{3}{4}x=33 \implies \frac{8}{4}x+\frac{3}{4}x=33 \implies \frac{11}{4}x=33 \implies x=33 	imes \frac{4}{11}=12$
## Step3: Substitute x to find y
$y=\frac{1}{4}(12)=3$

# Answer:
60. $\boldsymbol{(2, 2)}$
61. $\boldsymbol{(-2, 15)}$
62. $\boldsymbol{(7, -1)}$
63. $\boldsymbol{(2, 5)}$
64. $\boldsymbol{(2, 2)}$
65. $\boldsymbol{(12, 3)}$

QUESTION IMAGE

Question

Explanation:

For Questions 46-50 (Inequalities)

Question 46: $0 \geq -6 - 2x - 5y$

Step1: Rearrange to slope-intercept form

Step2: Graph the boundary line

Question 47: $y > |x|$

Step1: Identify the boundary graph

Step2: Determine shaded region

Question 48: $y \geq |x-3|$

Step1: Identify the boundary graph

Step2: Determine shaded region

Question 49

Step1: Calculate slope of boundary line

Step2: Write boundary line equation

Step3: Test point to find inequality sign

Question 50

Step1: Identify Quadrant I constraints

Step2: Determine valid inequalities

Question 51: $y=2x-4$

Step1: Check linearity

Step2: Graph key points

Question 52: $y=x^2+2$

Step1: Check linearity

Question 53: $y=3$

Step1: Check linearity

Step2: Graph the line

Question 54: $(0,9)$ and $(5,2)$

Step1: Calculate slope

Step2: Write slope-intercept form

Question 55: $(-5,-2)$ and $(7,7)$

Step1: Calculate slope

Step2: Solve for b

Question 56: $(0,0)$ and $(-8,-10)$

Step1: Calculate slope

Step2: Write slope-intercept form

Question 57: $(-1,-2)$ and $(1,4)$

Step1: Calculate slope

Step2: Solve for b

Question 58: $(2,2)$ and $(6,5)$

Step1: Calculate slope

Step2: Solve for b

Question 59: $(-3,2)$ and $(3,-1)$

Step1: Calculate slope

Step2: Solve for b

Answer:

For Questions 51-53 (Linear Functions)