Question

spiraling homework 7.2/7.3

1. a system of equations is represented by line h and j. a table representing some points on line h and the graph of line j

line h
x -16 -8 -4 12
y 7 1 -2 -14
are shown.
write a system in standard form for these two lines.
5.8 notes)

1. what value of x makes this equation true?

\\(\frac{2x - 7}{3} = \frac{52 - 4x}{-4}\\)

Explanation:

Problem 6: System of Equations for Lines h and j

Step 1: Find the equation of Line h

We have points on Line h: \((-16, 7)\), \((-8, 1)\), \((-4, -2)\), \((12, -14)\). Let's calculate the slope \(m\) using two points, say \((-16, 7)\) and \((-8, 1)\):

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 7}{-8 - (-16)} = \frac{-6}{8} = -\frac{3}{4} $$

Using point-slope form \(y - y_1 = m(x - x_1)\) with \((-8, 1)\):

$$ y - 1 = -\frac{3}{4}(x + 8) $$

Simplify to slope-intercept form:

$$ y - 1 = -\frac{3}{4}x - 6 \implies y = -\frac{3}{4}x - 5 $$

Multiply by 4 to eliminate fraction: \(4y = -3x - 20 \implies 3x + 4y = -20\) (standard form for Line h).

Step 2: Find the equation of Line j

From the graph, Line j passes through \((0, 0)\) and \((4, -4)\) (or other points). Slope \(m\):

$$ m = \frac{-4 - 0}{4 - 0} = -1 $$

Slope-intercept form: \(y = -x\) (since y-intercept is 0). Standard form: \(x + y = 0\).

Problem 8: Solve \(\frac{2x - 7}{3} = \frac{52 - 4x}{-4}\)

Step 1: Cross-multiply to eliminate denominators

$$ -4(2x - 7) = 3(52 - 4x) $$

Step 2: Distribute both sides

$$ -8x + 28 = 156 - 12x $$

Step 3: Add \(12x\) to both sides

$$ 4x + 28 = 156 $$

Step 4: Subtract 28 from both sides

$$ 4x = 128 $$

Step 5: Divide by 4

$$ x = 32 $$

Final Answers

Problem 6:

System of equations:
\(3x + 4y = -20\) (Line h)
\(x + y = 0\) (Line j)

Problem 8:

\(x = 32\)

Answer:

Problem 6: System of Equations for Lines h and j

Step 1: Find the equation of Line h

We have points on Line h: \((-16, 7)\), \((-8, 1)\), \((-4, -2)\), \((12, -14)\). Let's calculate the slope \(m\) using two points, say \((-16, 7)\) and \((-8, 1)\):

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 7}{-8 - (-16)} = \frac{-6}{8} = -\frac{3}{4} $$

Using point-slope form \(y - y_1 = m(x - x_1)\) with \((-8, 1)\):

$$ y - 1 = -\frac{3}{4}(x + 8) $$

Simplify to slope-intercept form:

$$ y - 1 = -\frac{3}{4}x - 6 \implies y = -\frac{3}{4}x - 5 $$

Multiply by 4 to eliminate fraction: \(4y = -3x - 20 \implies 3x + 4y = -20\) (standard form for Line h).

Step 2: Find the equation of Line j

From the graph, Line j passes through \((0, 0)\) and \((4, -4)\) (or other points). Slope \(m\):

$$ m = \frac{-4 - 0}{4 - 0} = -1 $$

Slope-intercept form: \(y = -x\) (since y-intercept is 0). Standard form: \(x + y = 0\).

Problem 8: Solve \(\frac{2x - 7}{3} = \frac{52 - 4x}{-4}\)

Step 1: Cross-multiply to eliminate denominators

$$ -4(2x - 7) = 3(52 - 4x) $$

Step 2: Distribute both sides

$$ -8x + 28 = 156 - 12x $$

Step 3: Add \(12x\) to both sides

$$ 4x + 28 = 156 $$

Step 4: Subtract 28 from both sides

$$ 4x = 128 $$

Step 5: Divide by 4

$$ x = 32 $$

Final Answers

Problem 6:

System of equations:
\(3x + 4y = -20\) (Line h)
\(x + y = 0\) (Line j)

Problem 8:

\(x = 32\)