Page 17 Essential Question Answer
To solve absolute value equations algebraically, first, isolate the absolute value expression on one side of the equation the same way you would isolate a variable.
Then use the rule: If x=a (where a is a positive number), then x=a OR x=−a.
To solve an absolute value equation, isolate the absolute value on one side of the equation. Then set its contents equal to both the positive and negative value of the number on the other side of the equation and solve both equations.
Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions
Page 17 Exercise 1 Answer
Given: The expression is |x+2|=3.
To find The value of x+2 that makes the equation true and d write it in an absolute form. Evaluate to get the answer.
We know that an absolute value equation can have either a positive or a negative value:
Set up the equations:
|x+2|=3:(x+2)=3…………….(1)
|x+2|=3:(x+2)=−3…………….(2)
The required solution is (x+2)=3 and(x+2)=−3.
Given: The expression is x+2=3.
To find The solution of the absolute value equation. Evaluate to get the answer.
Solve for x using Eq(1)and Eq(2):
1. |x+2|=3:(x+2)=3
x=3−2
x=1
2. |x+2|=3:(x+2)=−3
x=−3−2
x=−5
The required solutions are x=1 and x=−5.
We can use linear equations to solve absolute value problems by separating the equations into positive and negative components.
We can use linear equations to solve absolute value problems by separating the equations into positive and negative components.
Page 18 Exercise 3 Answer
Given: The absolute expression is x+2=3.
To find The values of absolute equations using a spreadsheet. Evaluate to get the answer.
We can solve the values for x+2 using a spreadsheet as follows:
The required solution is :
Given: The expression is x+2=3.
To compare: The solutions of the spreadsheet with the solutions in Explorations 1 and 2.
We observe that the value for(x+3)=3 and =-3 will be as follows:
The required answer is :
The absolute value equation is x+2=3.
We can solve absolute value equations using a spreadsheet by plugging in the values of x in the equation.
We can solve absolute value equations using a spreadsheet by plugging in the values of in the equation.
Page 18 Exercise 4 Answer
We can solve an absolute value equation using these steps:
Separate the absolute value equation to positive and negative components.
Set up these equations.
Solve for the values of the positive and negative components.
We separate the absolute value equation into positive and negative components and solve for these values.
Page 18 Exercise 5 Answer
We are given the absolute value equation: |f(x)|=c.
Solving the equation algebraically means solving two equations: f(x)=c and (x)=c. It is easy that we do not need graphics or tables, but we have to solve two equations.
Solving the equation graphically means drawing the number line and locating the point for which f(x)=0.
Then we locate the points c units from the point we determined. It is easy because we only solve one equation, but we have to draw a graph.
Solving the equation numerically means building a table of values and trying to find those values of x for which|f(x)|=c. It is easy because we don’t solve any equation and we don’t draw a graph, but we have to build a table and sometimes search among a big number of values.
On solving the equation algebraically we have to solve two equations. On solving equations graphically we only solve one equation, but we have to draw a graph. To solve the equation numerically we have to build a table and sometimes search among a big number of values.
Page 21 Exercise 4 Answer
Given: The expression is |d/3|=3.
To plot: The graph of solution on a number line. Evaluate to get the answer.
On solving the given equation :
solve for d: |d/3|=3
1. |d/3|=3: (d/3)=3
d=3.3
d=9
2. |d/3|=3: (d/3)=−3
d=−3⋅3
d=−9
Plot d=9 and d=−9
The required solution is d=9 and d=−9.
Page 21 Exercise 5 Answer
Given: The expression is 3 |2x+5|+10=37.
To plot: The graph of solution on a number line. Evaluate to get the answer.
On solving the given equation :
solve for x,
3|2x+5|+10=37
3(2x+5)=37−10 (distribute)
3.2x+3.5&=27 (simplify)
6x+15=27 (isolate x)
6x=27−15
6x=12
x=12/6
x=2
Solve for x:
3(2x+5)=−(37−10) (distribute)
3⋅2x+3⋅5=−2 (simplify )
6x+15=−27 (isolate x)
6x=−27−15
6x=−42
x=−42/6
x=−7
Plot x=2 and x=−7 on a number line :
Final answer The required solution is and.
Page 21 Exercise 6 Answer
Given: The expression is |20x|=|4x+16|.
To plot: The graph of solution on a number line. Evaluate to get the answer.
On solving the given equation :
solve for x:
|20x|=|4x+16|
20x=4x+16
20x−4x=16
16x=16
16x=16
x=16/16
x=1
Solve for x:
20x =−(4x+16) (distribute)
20x=−4x−16 (isolate x)
20x+4x=−16
24x=−16
24x=−16
x=−16/24
x=−2/3
Plot x=1 and x=−2/3:
The required solutions are x=1 and x=−2/3.
Page 21 Exercise 7 Answer
Given: The expression is |p+4|=|p−2|.
To plot: The graph of solution on the number line. Evaluate to get the answer.
On solving the equation :
Solve for p :
|p+4|=|p−2|
(p+4)=±(p−2)
(p+4)=(p−2) (isolate p)
p−p=−2−4 (simplify)
0=−6
It has no solution.
Solve for p:
(p+4)=−(p−2) (distribute)
p+4=−p+2 (isolate p)
p+p=2−4
2p=−2
p=−2/2
p=−1
Plot p=−1:
The required solution is p=−1.
Page 21 Exercise 8 Answer
Given: The expression is|4q+9|=|2q−1|.
To plot: The graph of solution on the number line. Evaluate to get the answer.
On solving the given equation :
solve for q:
|4q+9|=|2q−1|
(4q+9)=±(2q−1)
(4q+9)=(2q−1) (isolate q)
4q−2q=−1−9 (simplify )
2q=−10
q=−10/2
q=−5
solve for q:
(4q+9)=−(2q−1) (distribute)
(4q+9)=−2q+1 (isolate q)
4q+2q=1−9 (simplify)
6q=−8
q=−8/6
q=−4/3
Plot q=−5 and q=−4/3:
The required solutions are q=−5 and q=−4/3.
Page 21 Exercise 9 Answer
Given: The expression is |2x−7|=|2x+9|.
To plot: The graph of solution on the number line. Evaluate to get the answer.
On solving the equation :
solve for x,
|2x−7|=|2x+9|
(2x−7)=±(2x+9)
2x−7=2x+9 (isolate x)
2x−2x=9+7 (simplify )
0=16
It has no solution.
Solve for x:
2x−7=−(2x+9) (distribute)
2x−7=−2x−9 (isolate)
2x+2x=−9+7 (simplify)
4x=−2
x=−2/4
x=−1/2
Plot x=−1/2:
The required solution is x=−1/2.
