Algebra 1
Algebra 1
1st Edition
ISBN: 9780078884801
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Differential Equations
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Chapter 3.1, Problem 25PPS
To determine

Graph the equation by using x and y − intercept.

Expert Solution & Answer
Check Mark

Answer to Problem 25PPS

x and y intercepts are x= 5 and y = − 1

Explanation of Solution

Given:

The equation: x=5y+5

Concept Used:

The  x -intercepts are where the graph crosses the  x -axis, and the  y -intercepts are where the graph crosses the  y -axis.

Then, algebraically,

  • an  x -intercept is a point on the graph where  y  is zero, and
  • a  y -intercept is a point on the graph where  x  is zero.

More specifically,

  • an  x -intercept is a point in the equation where the  y -value is zero, and
  • a  y -intercept is a point in the equation where the  x -value is zero.

Calculation:

The equation: x=5y+5

Find the x − intercept, plug in y = 0 in the equation.

  x=5y+5x=5  ·  0 +  5   [  plug  in  y  =0  ]x=5 x − Intercept: x= 5

Find the y − intercept, plug in x = 0 in the equation

  x=5y+50=5  y +  5   [  plug  in  x  =0  ]0=5y+5−5=5y+5−5  [ Subtract    from  both  sides ]  −5=5y−55  =  5y5   [  Divide  each  side  by  5  ]y=  −1 y − Intercept: y = − 1

Graph the line x=5y+5 with x and y intercepts points are (5, 0) and (0, − 1)

  Algebra 1, Chapter 3.1, Problem 25PPS

Thus, x and y intercepts are x = 5 and y = − 1

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Chapter 3 Solutions

Algebra 1

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