body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

4. Graphing Linear Equations in Standard Form

Finish the assignment

Continue to next subchapter

Exercise 29 Page 134

Find the intercepts by substituting x=0 and y=0 into the equation.

A

Practice makes perfect

We are given the following equation. 5x+3y=30 We want to match the equation with one of the given graphs. Note that each graph has a different set of x- and y-intercepts. By finding these one at a time, we can identify which graph belongs to which equation.

x-intercept

The x-intercept occurs when y=0. By substituting 0 for y into our equation and solving for x, we can find the x-intercept. Let's do it!

5x+3y=30

y= 0

5x+3( 0)=30

▼

Solve for x

Zero Property of Multiplication

5x+0=30

Identity Property of Addition

5x=30

.LHS /5.=.RHS /5.

x=6

We obtained that the x-intercept is at the point (6,0).

y-intercept

The y-intercept occurs when x=0.To find the y-intercept, we substitute 0 for x and solve for y. Let's do it!

5x+3y=30

x= 0

5( 0)+3y=30

▼

Solve for y

Zero Property of Multiplication

0+3y=30

Identity Property of Addition

3y=30

.LHS /3.=.RHS /3.

y=10

We obtained that the y-intercept is at the point (0,10).

Graph

Now that we know the intercepts, we can draw the graph of this equation.

We can see that this graph corresponds to option A.

Subchapter links

Communicate Your Answer p.129

Monitoring Progress p.130-132

Exercises p.133-134