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Practice Problem #1

Put the equation into standard form. y = −5x + 11

Start by moving the x term to the left side of the equation.

  • (Fill in the blank) ____ + y = 11

Hint: Since x is neither negative or a fraction, we are now done.

Practice Problem #2

Put the equation into standard form. y equals negative one-half x minus 4

Start by moving the x term to the left side of the equation

  • blank over a blank times x plus y equals negative 4

Since the coefficient of the x term is a fraction, we need to eliminate the fraction.

  • ( (Fill in the blank)____)(one-half x plus y) = ((Fill in the blank)____)(−4)
  • Simplifiy.
  • x + (Fill in the blank) ____ = (Fill in the blank) ____

Practice Problem #3

Put the equation into standard form. y equals two-sevenths x plus 1

Start by moving the x term to the left side of the equation.

  • a blank over a blank times x plus y = 1
  • Since the coefficient of the x term is a fraction, we need to eliminate the fraction.
  • ( (Fill in the blank) ____ )(negative 2 over 7 x plus y) = ( (Fill in the blank) ____ )(1)
  • Simplify.
  • (Fill in the blank) ____ + (Fill in the blank) ____ = (Fill in the blank) ____
  • We need to make sure x has a positive coefficient. Switch the signs of all terms to complete this equation.
  • (Fill in the blank) ____ 2x (Fill in the blank) ____ 7y = (Fill in the blank) ____ 7

Practice Problem #4 Part 1

Write the equation of the line parallel to the line 5x − 2y = 7 with a y-intercept at (0, 4). Write your answer in standard form.

Before we get started we need to review. When lines are parallel, their slopes are:

  1. equal
  2. negative reciprocals

Practice Problem #4 Part 2

Good, so we know the slopes of our lines are equal. OK, we have an intercept, but we need a slope. Let's find the slope of the given line by changing it from standard form to slope intercept form.

  • 5x − 2y = 7
  • (Fill in the blank) ____ (Fill in the blank) = (Fill in the blank) ____ (Fill in the blank) + 7 Move the x term to the right side
  • The y term should have a coefficient of 1. Divide both sides by −2 to get this coefficient.
  • y equals a blank over a blank times x minus a blank over a blank
  • What is the slope of this line?

Practice Problem #4 Part 3

Now we know the slope of the line with the equation 5x − 2y = 7 is five-halves. We can use this slope to write the next equation in slope intercept form.

  • y = mx + b Substitute in the values for slope and y-intercept
  • y equals a blank over a blank times x plus 4
  • Ok, now we know simply need to change this to standard form. Start by moving the x term to the left side of the equation.
  • negative a blank over a blank times x plus y equals to 4
  • Since the coefficient of the x term is a fraction, we need to eliminate the fraction.
  • ( (Fill in the blank) ____ )(negative five-halves x plus 1) = ( (Fill in the blank) ____ )(4)

  • Simplify.

    (Fill in the blank) ____ + (Fill in the blank) ____ = (Fill in the blank) ____
  • We need to make sure x has a positive coefficient. Switch the signs of all terms to complete this equation.
  • (Fill in the blank) ____ − (Fill in the blank) ____ = (Fill in the blank) ____

Practice Problem #5 Part 1

Write the equation of the line perpendicular to the line 5x + 7y = 12 and passes through the point (5, 2). Write your answer in standard form.

Before we get started we need to review. When lines are perpendicular, their slopes are:

  1. negative reciprocals
  2. equal

Practice Problem #5 Part 2

OK, we have a point, but need a slope. Let's find the slope of the given line by changing it from standard form to slope intercept form.

  • 5x + 7y = 12

  • Move the x term to the right side

    7y = (Fill in the blank) ____ + 12
  • The y term should have a coefficient of 1. Divide both sides by 7 to get this coefficient.
  • y equals negative a blank over a blank times x plus a blank over a blank
  • What is the slope of this line?

Practice Problem #5 Part 3

Now we know the slope of the line with the equation 5x + 7y = 12 is negative five-sevenths. If the slope of the perpendicular line is the negative reciprocal, the perpendicular line's slope is:

For our second line, we now have a slope (seven-fifths) and a point, (5, 2). We can use the point slope form to find this line's equation.

  • y − y = m(x − x)

Substitute in the points given: y − (Fill in the blank) ____ = seven-fifths(x − (Fill in the blank) ____ )

Use the distributive property to eliminate the parentheses: y − 2= seven-fifthsx − (Fill in the blank) ____

Move all constant values to the right side: y = seven-fifthsx − (Fill in the blank) ____ (Fill in the blank)

Practice Problem #5 Part 4

Now we have point slope form of the line. The last step is to change it to the standard form.

  • y = seven-fifthsx − 5
  • Start by moving the x term to the left side
  • −seven-fifthsx + y = −5
  • Eliminate the fraction in the x coefficient by multiplying both sides by the denominator:
  • ( (blank) (Fill in the blank) ____ )(−seven-fifthsx + y = ( (blank) (blank) (Fill in the blank) ____ )(−5)
  • Simplify
  • (Fill in the blank) ____ (Fill in the blank) + (Fill in the blank) ____ (Fill in the blank) = (Fill in the blank) ____ (Fill in the blank)
  • Since the coefficient of the x term should be positive, switch the signs for all terms.
  • (Fill in the blank)____ (Fill in the blank) + (Fill in the blank) ____ = (Fill in the blank) ____ (Fill in the blank)