Solve the equation. check for extraneous solutions.

9|9 – 8x| = 2x + 3

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Solve the equation. check for extraneous solutions.

9|9 – 8x| = 2x + 3

81-72x=2x+3

81-3=2x+72x

78=74x

x=74/78

x=37/39 or 0.948...

To Check for extraneous solutions, you have to plug in the answers to the original equation. If it doesn’t work then it is extraneous and it is no solution

81-72=2X+3

-3 -3

78-72=2X

6=2X

2/6=3

3=X

39/37 and 6/5

Step-by-step explanation:

To start, let's assume that 9-8x is positive (we will do this to get rid of the absolute value, but then check later). We get 9(9 - 8x) = 2x + 3, or 81 - 72x = 2x + 3, so 74x = 78 and x = 39/37. If we assume 9-8x is negative, the |9-8x| will turn into 8x-9, so we get 9(8x - 9) = 2x + 3, or 72x - 81 = 2x + 3 , so 70x = 84 and x = 6/5. If we plug the values back in, 9-8(39/37) needs to be positive, which works, and 9-8(6/5) needs to be negative, and this works, so the answers are 39/37 and 6/5.

The given equation is : 9|9-8x| = 2x + 3. Removing the modulus sign to get, 9(9-8x) = +2x+3 or 9(9-8x) = -2x-3 . Case 1: 9(9-8x) = +2x+3 or 81 -72x = 2x +3 or 78 = 74x or x = 78/74. Case 2:9(9-8x) = -2x-3 or 81 - 72x =- 2x -3 or 70x = 84 or x = 84/70. Thus, the solutions of x are 78/74 and 84/70. No, these are not extraneous solutions, as the answers are satisfying the equation.

The solution of 9|9 – 8x| = 2x + 3 is 39/37 or 6/5

Further explanation

We need to recall the following definitions to solve absolute equation.

[tex]| x - a | = \sqrt { (x - a)^2 }[/tex]

If | x | > a then x > a or x < -a

If | x | < a then -a < x < a

Let us tackle the problem.

This problem is about linear absolute equation.

Given:

[tex]9 | 9 - 8x | = 2x + 3[/tex]

If x ≤ 9/8 , then :

[tex]9 ( 9 - 8x ) = 2x + 3[/tex]

[tex]81 - 72x = 2x + 3[/tex]

[tex]81 - 3 = 72x + 2x[/tex]

[tex]78 = 74x[/tex]

[tex]x = 78 / 74[/tex]

[tex]x = \boxed {\frac{39}{37}}[/tex]

If x > 9/8 , then :

[tex]-9 ( 9 - 8x ) = 2x + 3[/tex]

[tex]-81 + 72x = 2x + 3[/tex]

[tex]-81 - 3 = -72x + 2x[/tex]

[tex]-84 = -70x[/tex]

[tex]x = -84 / -70[/tex]

[tex]x = \boxed {\frac{6}{5}}[/tex]

Learn moreInfinite Number of Solutions : System of Equations : System of Linear equations : Answer details

Grade: High School

Subject: Mathematics

Chapter: Absolute Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point , Absolute

[tex]Solve the equation. check for extraneous solutions. 9|9 – 8x| = 2x + 3[/tex]

[tex]9-8x\geq0\ \ \ |subtract\ 9\ from\ both\ sides\\\\-8x\geq-9\ \ \ \ |change\ signs\\\\8x\leq9\ \ \ \ |divide\ both\ sides\ by\ 8\\\\x\leq9\\\\ |9-8x|=\left\{\begin{array}{ccc}9-8x&for\ x\leq9\\8x-9&for\ x \ \textgreater \ 9\end{array}\right[/tex]

[tex]1^o\ x\in(-\infty;\ 9]\to|9-8x|=9-8x\\\\9(9-8x)=2x+3\\81-72x=2x+3\ \ \ \ |subtract\ 81\ from\ both\ sides\\-72x=2x-78\ \ \ \ |subtract\ 2x\ from\ both\ sides\\-74x=-78\ \ \ \ |divide\ both\ sides\ by\ (-74)\\\\x=\dfrac{-78:2}{-74:2}\\\\\boxed{x=\frac{39}{37}}\in(-\infty;\ 9][/tex]

[tex]2^o\ x\in(9;\ \infty)\to|9-8x|=8x-9\\\\9(8x-9)=2x+3\\72x-81=2x+3\ \ \ \ |add\ 81\ to\ both\ sides\\72x=2x+84\ \ \ \ |subtract\ 2x\ from\ both\ sides\\70x=84\ \ \ \ \ \ |divide\ both\ sides\ by\ 70\\\\x=\dfrac{84:14}{70:14}\\\\x=\dfrac{6}{5}\notin(9;\ \infty)\\\\\\\boxed{x=\frac{39}{37}}[/tex]

9|9-8x|=2x+3

As there is mod, there would be two solutions, one for +ve value and other for -ve valueIt will lead into either

1) 9(9-8x)=2x+3 OR

2) 9(9-8x)=-(2x+3)

Now,

1)9(9-8x)=2x+3

81-72x=2x+3

74x=78

x=39/37

2)9(9-8x)=-(2x+3)

81-72x=-2x-3

70x=84

x=6/5

So, x would be either 39/37 or 6/5

9|9-8x|=2x+3 divide both sides by 9

l 9 - 8x l = [2x + 3] / 9 we have two equations

9 - 8x = [2x + 3] / 9 and 9 - 8x = - [2x + 3] / 9

multiply both sides by 9

81 - 72x = 2x + 3 81 - 72x = -2x - 3

get the x's on one side and constants on the other

78 = 74x 84 = 70x

divide by 74 divide by 70

78/74 = x = 39/37 84/70 = x = 6/5

81-72x = 2x + 3

-74x = -84

x = 84/74= 42/37

-81 + 72x = 2x + 3

70x = 84

x = 84/70= 42/35= 6/5

12

expolntion:

12

=-11

13=12