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Question 5 (Essay Worth 10 points)
(01.03 MC)
Solve for x and show your steps. Is the solution extraneous? Check your work to show how you determined if the solution is extraneous or not.

Question 5 Essay Worth 10 points 0103 MC Solve for x and show your steps Is the solution extraneous Check your work to show how you determined if the solution i class=

Respuesta :

semsee45

Answer:

x = 4

Step-by-step explanation:

Given:

[tex]\sqrt{11x+5}=7[/tex]

Square both sides:

[tex]\implies \left(\sqrt{11x+5}\right)^2=7^2[/tex]

[tex]\implies 11x+5=49[/tex]

Subtract 5 from both sides:

[tex]\implies 11x+5-5=49-5[/tex]

[tex]\implies 11x=44[/tex]

Divide both sides by 11:

[tex]\implies 11x \div 11=44 \div 11[/tex]

[tex]\implies x=4[/tex]

fieryanswererft

Answer:

 x = 4

Explanation:

[tex]\sf \rightarrow \sqrt{11x+ 5} = 7[/tex]

square both sides

[tex]\sf \rightarrow 11x+ 5 = 7^2[/tex]

simplify the following

[tex]\sf \rightarrow 11x+ 5 = 49[/tex]

subtract 5 from both sides

[tex]\sf \rightarrow 11x+ 5 -5 = 49-5[/tex]

simplify the following

[tex]\sf \rightarrow 11x = 44[/tex]

divide both sides by 11

[tex]\sf \rightarrow \dfrac{ 11x}{11} = \dfrac{44}{11}[/tex]

simplify the following

[tex]\sf \rightarrow x=4[/tex]

Check for Solution:

[tex]\sf \rightarrow \sqrt{11(4)+ 5} = 7[/tex]

[tex]\sf \rightarrow \sqrt{44+ 5} = 7[/tex]

[tex]\sf \rightarrow \sqrt{49} = 7[/tex]

[tex]\sf \rightarrow 7 = 7[/tex]

Hence proved the answer as L.H.S = R.H.S.