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practice example 1 write an equation of a line in slope - intercept for…

Question

practice
example 1
write an equation of a line in slope - intercept form with the given slope and y - intercept.

  1. slope: 5, y - intercept: - 3
  2. slope: - 2, y - intercept: 7
  3. slope: - 6, y - intercept: - 2
  4. slope: 7, y - intercept: 1
  5. slope: 3, y - intercept: 2
  6. slope: - 4, y - intercept: - 9
  7. slope: 1, y - intercept: - 12
  8. slope: 0, y - intercept: 8

example 2
write each equation in slope - intercept form.

  1. - 10x + 2y = 12
  2. 4y + 12x = 16
  3. - 5x + 15y = - 30
  4. 6x - 3y = - 18
  5. - 2x - 8y = 24
  6. - 4x - 10y = - 7

Explanation:

Response

Let's solve problem 9: \(-10x + 2y = 12\) (writing in slope - intercept form \(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept)

Step 1: Isolate the \(y\) - term

We start with the equation \(-10x + 2y=12\). First, we add \(10x\) to both sides of the equation to get the \(y\) - term by itself.
\(-10x + 2y+10x=12 + 10x\)
Simplifying the left - hand side, \(-10x+10x = 0\), so we have \(2y=10x + 12\)

Step 2: Solve for \(y\)

Now, we divide every term in the equation \(2y = 10x+12\) by 2 to solve for \(y\).
\(y=\frac{10x}{2}+\frac{12}{2}\)
Simplifying the fractions, \(\frac{10x}{2}=5x\) and \(\frac{12}{2} = 6\). So \(y = 5x+6\)

Let's solve problem 10: \(4y+12x = 16\)

Step 1: Isolate the \(y\) - term

Start with \(4y + 12x=16\). Subtract \(12x\) from both sides of the equation.
\(4y+12x-12x=16 - 12x\)
Simplifying the left - hand side, \(12x-12x = 0\), so \(4y=-12x + 16\)

Step 2: Solve for \(y\)

Divide every term in the equation \(4y=-12x + 16\) by 4.
\(y=\frac{-12x}{4}+\frac{16}{4}\)
Simplifying the fractions, \(\frac{-12x}{4}=-3x\) and \(\frac{16}{4}=4\). So \(y=-3x + 4\)

Let's solve problem 11: \(-5x + 15y=-30\)

Step 1: Isolate the \(y\) - term

Start with \(-5x + 15y=-30\). Add \(5x\) to both sides of the equation.
\(-5x + 15y+5x=-30 + 5x\)
Simplifying the left - hand side, \(-5x + 5x=0\), so \(15y=5x-30\)

Step 2: Solve for \(y\)

Divide every term in the equation \(15y = 5x-30\) by 15.
\(y=\frac{5x}{15}-\frac{30}{15}\)
Simplifying the fractions, \(\frac{5x}{15}=\frac{1}{3}x\) and \(\frac{30}{15} = 2\). So \(y=\frac{1}{3}x-2\)

Let's solve problem 12: \(6x-3y=-18\)

Answer:

s:

Example 1 (Write equation in slope - intercept form):
  1. \(y = 5x-3\)
  2. \(y=-2x + 7\)
  3. \(y=-6x-2\)
  4. \(y=7x + 1\)
  5. \(y=3x + 2\)
  6. \(y=-4x-9\)
  7. \(y=x-12\)
  8. \(y = 8\)
Example 2 (Write equation in slope - intercept form):
  1. \(y = 5x+6\)
  2. \(y=-3x + 4\)
  3. \(y=\frac{1}{3}x-2\)
  4. \(y = 2x+6\)
  5. \(y=-\frac{1}{4}x-3\)
  6. \(y=-\frac{2}{5}x+\frac{7}{10}\)