pre-algebra 10-3 solving equations with variables on both sides
TRANSCRIPT
Pre-Algebra
10-3
Solving Equations with Variables on Both Sides
Warm UpSolve.
1. 2x + 9x – 3x + 8 = 16
2. –4 = 6x + 22 – 4x
3. + = 5
4. – = 3
x = 1
x = -13
x = 3427
x7 7
1
9x16
2x4
18
x = 50
Learn to solve equations with variables on both sides of the equal sign.
Some problems produce equations that have variables on both sides of the equal sign.
Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.
Solve.
A. 4x + 6 = x
Example: Solving Equations with Variables on Both Sides
4x + 6 = x– 4x – 4x
6 = –3x
Subtract 4x from both sides.
Divide both sides by –3.
–2 = x
6–3
–3x–3=
Solve.
B. 9b – 6 = 5b + 18
Example: Solving Equations with Variables on Both Sides
9b – 6 = 5b + 18– 5b – 5b
4b – 6 = 18
4b 4
24 4 =
Subtract 5b from both sides.
Divide both sides by 4.
b = 6
+ 6 + 6
4b = 24Add 6 to both sides.
Solve.
C. 9w + 3 = 5w + 7 + 4w
Example: Solving Equations with Variables on Both Sides
9w + 3 = 5w + 7 + 4w
3 ≠ 7
9w + 3 = 9w + 7 Combine like terms.
– 9w – 9w Subtract 9w from both sides.
No solution. There is no number that can be substituted for the variable w to make the equation true.
Solve.
A. 5x + 8 = x
Try This
5x + 8 = x– 5x – 5x
8 = –4x
Subtract 4x from both sides.
Divide both sides by –4.
–2 = x
8–4
–4x–4=
Solve.
B. 3b – 2 = 2b + 123b – 2 = 2b + 12
– 2b – 2b
b – 2 = 12
Subtract 2b from both sides.
+ 2 + 2
b = 14Add 2 to both sides.
Try This
Solve.
C. 3w + 1 = 10w + 8 – 7w3w + 1 = 10w + 8 – 7w
1 ≠ 8
3w + 1 = 3w + 8 Combine like terms.
– 3w – 3w Subtract 3w from both sides.
No solution. There is no number that can be substituted for the variable w to make the equation true.
Try This
To solve multistep equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.
Solve.
A. 10z – 15 – 4z = 8 – 2z - 15
Example: Solving Multistep Equations with Variables on Both Sides
10z – 15 – 4z = 8 – 2z – 15
+ 15 +15
6z – 15 = –2z – 7 Combine like terms.+ 2z + 2z Add 2z to both sides.
8z – 15 = – 7
8z = 8
z = 1
Add 15 to both sides.
Divide both sides by 8.8z 88 8=
B.
Example: Solving Multistep Equations with Variables on Both Sides
Multiply by the LCD.
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14 Combine like terms.
y5
34
3y5
710
+ – = y –
y5
34
3y5
710
+ – = y –
20( ) = 20( )y5
34
3y5
710
+ – y –
20( ) + 20( ) – 20( )= 20(y) – 20( )y5
3y5
34
710
Example Continued
Add 14 to both sides.
–15 = 4y – 14
–1 = 4y
+ 14 + 14
–1 4
4y4 = Divide both sides by 4.
-14 = y
16y – 15 = 20y – 14
– 16y – 16y Subtract 16y from both sides.
Solve.
A. 12z – 12 – 4z = 6 – 2z + 32
Try This
12z – 12 – 4z = 6 – 2z + 32
+ 12 +12
8z – 12 = –2z + 38 Combine like terms.+ 2z + 2z Add 2z to both sides.
10z – 12 = + 38
10z = 50
z = 5
Add 12 to both sides.
Divide both sides by 10.10z 5010 10=
B.
Multiply by the LCD.
6y + 20y + 18 = 24y – 18
26y + 18 = 24y – 18 Combine like terms.
y4
34
5y6
68
+ + = y –
y4
34
5y6
68
+ + = y –
24( ) = 24( )y4
34
5y6
68
+ + y –
24( ) + 24( )+ 24( )= 24(y) – 24( )y4
5y6
34
68
Try This
Subtract 18 from both sides.
2y + 18 = – 18
2y = –36
– 18 – 18
–36 2
2y2 = Divide both sides by 2.
y = –18
26y + 18 = 24y – 18
– 24y – 24y Subtract 24y from both sides.
Try This Continued
Example: Consumer Application
Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?
Example Continued
First solve for the price of one doughnut.
1.25 + 2d = 0.50 + 5dLet d represent the price of one doughnut.
– 2d – 2d
1.25 = 0.50 + 3dSubtract 2d from both sides.
– 0.50 – 0.50Subtract 0.50 from both sides.
0.75 = 3d
0.753
3d3= Divide both sides by 3.
0.25 = d The price of one doughnut is $0.25.
Example Continued
Now find the amount of money Jamie spends each morning.
1.25 + 2d Choose one of the original expressions.
Jamie spends $1.75 each morning.
1.25 + 2(0.25) = 1.75
0.25n0.25
1.75 0.25 =
Let n represent the number of doughnuts.
Find the number of doughnuts Jamie buys on Tuesday.
0.25n = 1.75
n = 7; Jamie bought 7 doughnuts on Tuesday.
Divide both sides by 0.25.
Try This
Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?
Try This Continued
First solve for distance around the track.
2x + 4 = 4x + 2Let x represent the distance around the track.
– 2x – 2x
4 = 2x + 2Subtract 2x from both sides.
– 2 – 2 Subtract 2 from both sides.
2 = 2x
22
2x2= Divide both sides by 2.
1 = x The track is 1 mile around.
Try This Continued
Now find the total distance Helene walks each day.
2x + 4 Choose one of the original expressions.
Helene walks 6 miles each day.2(1) + 4 = 6
Let n represent the number of 1-mile laps.
Find the number of laps Helene walks on Saturdays.
1n = 6
Helene walks 6 laps on Saturdays.
n = 6
Lesson Quiz
Solve.
1. 4x + 16 = 2x
2. 8x – 3 = 15 + 5x
3. 2(3x + 11) = 6x + 4
4. x = x – 9
5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each?
x = 6
x = –8
no solution
x = 3614
12
An orange has 45 calories. An apple has 75 calories.