Download Standard Form

ALGEBRA 1 LESSON 6-3
Standard Form
(For help, go to Lessons 2-3 and 2-6.)
Solve each equation for y.
1. 3x + y = 5
2. y – 2x = 10
3. x – y = 6
4. 20x + 4y = 8
5. 9y + 3x = 1
6. 5y – 2x = 4
Clear each equation of decimals.
7. 6.25x + 8.5 = 7.75
8. 0.4 = 0.2x – 5
9. 0.9 – 0.222x = 1
8-5
ALGEBRA 1 LESSON 6-3
Standard Form
Solutions
1.
3x + y = 5
3x – 3x + y = 5 – 3x
y = –3x + 5
3. x – y = 6
x=6+y
x–6=y
y=x–6
5. 9y + 3x = 1
9y = –3x + 1
y = –3x+ 1
9
1
y=– x+ 1
3
9
2. y – 2x = 10
y – 2x + 2x = 10 + 2x
y = 2x + 10
4. 20x + 4y = 8
4y = –20x + 8
y = –20x + 8
4
y = –5x + 2
6. 5y – 2x = 4
5y = 2x + 4
y = 2x+ 4
5
2
y = x +4
5
5
8-5
7. Multiply each term by 100:
100(6.25x)+100(8.5) = 100(7.75)
Simplify:
625x + 850 = 775
8. Multiply each term by 10:
10(0.4) = 10(0.2x) – 10(5)
Simplify:
4 = 2x – 50
9.
Multiply each term
by 1000:
1000(0.9)–1000(0.222x)=1000(1)
Simplify:
900 – 222x = 1000
ALGEBRA 1 LESSON 6-3
Standard Form
Find the x- and y-intercepts of 2x + 5y = 6.
Step 1
To find the x-intercept,
substitute 0 for y and
solve for x.
Step 2
To find the y-intercept,
substitute 0 for x and
solve for y.
2x + 5y = 6
2x + 5y = 6
2x + 5(0) = 6
2(0) + 5y = 6
2x = 6
5y = 6
x=3
y=
The x-intercept is 3.
6
5
The y-intercept is
8-5
6
.
5
ALGEBRA 1 LESSON 6-3
Standard Form
Graph 3x + 5y = 15 using intercepts.
Step 1
Find the intercepts.
Step 2
Plot (5, 0) and (0, 3).
Draw a line through the points.
3x + 5y = 15
3x + 5(0) = 15
3x = 15
Substitute 0 for y.
Solve for x.
x=5
3x + 5y = 15
3(0) + 5y = 15
5y = 15
Substitute 0 for x.
Solve for y.
y=3
8-5
ALGEBRA 1 LESSON 6-3
Standard Form
a. Graph y = 4
b. Graph x = –3.
0 • x + 1 • y = 4 Write in standard form.
For all values of x, y = 4.
1 • x + 0 • y = –3 Write in standard form.
For all values of y, x = –3.
8-5
ALGEBRA 1 LESSON 6-3
Standard Form
Write y = 2 x + 6 in standard form using integers.
3
y= 2x+6
3
3y = 3( 2 x + 6 )
Multiply each side by 3.
3y = 2x + 18
Use the Distributive Property.
3
–2x + 3y = 18
Subtract 2x from each side.
The equation in standard form is –2x + 3y = 18.
8-5
ALGEBRA 1 LESSON 6-3
Standard Form
Write an equation in standard form to find the number of
hours you would need to work at each job to make a total of $130.
Job
Mowing
lawns
Delivering
newspapers
Amount Paid
per hour
Define: Let x = the hours mowing lawns.
Let y = the hours delivering
newspapers.
$12
$5
Relate: $12 per h plus
mowing
12 x
Write:
+
$5 per h equals $130
delivering
5 y
=
The equation standard form is 12x + 5y = 130.
8-5
130
ALGEBRA 1 LESSON 6-3
Standard Form
Find each x- and y-intercepts of each equation.
2. –4x – 3y = 9
1. 3x + y = 12
x = – 9 , y = –3
x = 4, y = 12
4
3. Graph 2x – y = 6 using x- and y-intercepts.
For each equation, tell whether its graph is
horizontal or vertical.
4. y = 3
horizontal
5. x = –8
vertical
6. Write y = 5 x – 3 in standard form using integers.
2
–5x + 2y = –6 or 5x – 2y = 6
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
(For help, go to Lessons 6-1 and 1-7.)
Find the rate of change of the data in each table.
1.
x
y
2
5
8
11
4
–2
–8
–14
2.
x
y
–3
–1
1
3
–5
–4
–3
–2
Simplify each expression.
4. –3(x – 5)
5. 5(x + 2)
6. – 4 (x – 6)
9
8-5
3.
x
y
10
4
7.5
–1
5
–6
2.5
–11
Point-Slope Form and Writing
Linear Equations
Solutions
ALGEBRA 1 LESSON 6-4
4 – (–2)
6
1. Use points (2, 4) and (5, –2). rate of change = 2 – 5 = –3 = –2
2. Use points (–3, –5) and (–1, –4). rate of change =
–5 – (–4) –5 + 4
1
=
= –1 =
–3 – (–1) –3 + 1
–2 2
3. Use points (10, 4) and (7.5, –1). rate of change =
4 – (–1 )
4+1
5
=
=
=2
10 – 7.5
2.5
2.5
4. –3(x – 5) = –3x – (–3)(5) = –3x + 15
5. 5(x + 2) = 5x + 5(2) = 5x + 10
6. – 4 (x – 6) = – 4 x – (– 4 )(6) = – 4 x + 8
9
9
9
9
3
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
Graph the equation y – 2 = 1 (x – 1).
3
The equation of a line that passes
through (1, 2) with slope 1 .
3
Start at (1, 2). Using the slope, go
up 1 unit and right 3 units to (4, 3).
Draw a line through the two points.
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
Write the equation of the line with slope –2 that passes
through the point (3, –3).
y – y1 = m(x – x1)
y – (–3) = –2(x – 3)
y + 3 = –2(x – 3)
Substitute (3, –3) for (x1, y1) and –
2 for m.
Simplify the grouping symbols.
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
Write equations for the line in point-slope form and in
slope-intercept form.
Step 1
Find the slope.
y 2 – y1
x 2 – x1
=m
4–3
–1 – 2
=–
The slope is – 1 .
3
8-5
1
3
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
(continued)
Step 2
Use either point to write
the the equation in
point-slope form.
Step 3
Rewrite the equation
from Step 2 in slope–
intercept form.
y – 4 = – 1 (x + 1)
Use (–1, 4).
3
y–4=–1x–1
3
3
y = – 1 x + 32
3
3
y – y1 = m(x – x1)
y – 4 = – 1 (x + 1)
3
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
Is the relationship shown by the data linear? If so, model the
data with an equation.
Step 1 Find the rate of change for
consecutive ordered pairs.
–1(
–3(
–2(
x
y
3
2
–1
–3
6
4
–2
–6
) –2
–2
=2
–1
) –6
) –4
–6
=2
–3
–2
=2
–1
The relationship is linear.
The rate of change is 2.
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
(continued)
Step 2
Use the slope 2 and a point (2,4) to write an equation.
y – y1 = m(x – x1) Use the point-slope form.
y – 4 = 2(x – 2)
Substitute (2, 4) for (x1, y1) and 2 for m.
8-5
Point-Slope Form and Writing
Linear Equations
ALGEBRA 1 LESSON 6-4
Is the relationship shown by the data linear? If so, model the
data with an equation.
Step 1 Find the rate of change for
consecutive ordered pairs.
1(
2(
1(
x
–2
–1
1
2
y
–2
–1
0
1
)1
1
1
=1
)1
)1
–1
2
=/ 1
–1
1
=1
–
The relationship is not linear.
8-5
Point-Slope Form and Writing
1. Graph the equation
y + 1 = –(x – Equations
3).
Linear
ALGEBRA 1 LESSON 6-4
2. Write an equation of the line with
slope – 2 that passes through
3
the point (0, 4).
y – 4 = – 2 (x – 0), or y = – 2 x + 4
3
3
3. Write an equation for the line that
passes through (3, –5) and (–2, 1) in
Point-Slope form and Slope-Intercept
form.
y + 5 = – 6 (x – 3); y = – 6 x – 7
5
5
5
4. Is the relationship shown by the data
linear? If so, model that data with an
equation.
2
yes; y + 3 = 5 (x – 0)
8-5
x
–10
0
5
20
y
–7
–3
–1
5