We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
283
3
avatar+569 

1.  (x + 7)/2 = 5x/3

 

2. 9/(- 3x) = 12/(x + 10)

 

3.  (j + 4)/j + 2 = 2 - 1/j

 

4. 8/q + 2 = q+4/q-1

 

 

 

5.

Heather wrote this solution to solve for x.

Which statement is true?

 

 

 

 

A. −1 is an extraneous solution because it creates a 0 in the denominator.

B. 5 is an extraneous solution because it creates a 0 in the denominator.

C.There are no extraneous solutions.

D.Both ​−1​ and 5 are extraneous solutions because they create a 0 in the denominator.

 Oct 19, 2018

Best Answer 

 #1
avatar+322 
+3

1)\(\frac{ (x + 7)}{2 }=\frac{5x}{3}\) <=> \(3 (x + 7) = 2(5x)\) <=> \(3x + 21 = 10x \) <=> \(7x=21\) <=> \(x=21/7\) <=> \(x=3\)

2)\(\frac{ 9}{(- 3x)}=\frac{12}{(x + 10)}\) <=> \( (x+10)9 = 12(-3x)\)<=> \(9x + 90 = -36x\)<=>\(-45x=90 \) <=> \(x=-2\)

3) \(\frac{ (j + 4)}{j}+2 = 2 -\frac{1}{j}\),\(j≠ 0\)<=> \(\frac{(j+4)}{j} = -\frac{1}{j}\)<=> \(j+4=-1\) <=> \(j=-5\)

4) \(\frac{8}{q }+2=\frac{(q+4)}{(q-1)}\),\(q≠0,q≠ 1\)<=> \(\frac{8+2q}{q }=\frac{(q+4)}{(q-1)}\)<=>\((q-1)(8+2q)=(q+4)(q)\)<=>\(8q+2q^2-8-2q=q^2+4q\) <=> \(q^2+2q-8=0\)<=>\((q-2)(q+4)=0 \)<=> \(q=2,q=-4\)

5) Answer \(B\) Because \(x-5\) is in the denominator and NOT \(x+1\)

 

Hope it helps! 

 Oct 19, 2018
 #1
avatar+322 
+3
Best Answer

1)\(\frac{ (x + 7)}{2 }=\frac{5x}{3}\) <=> \(3 (x + 7) = 2(5x)\) <=> \(3x + 21 = 10x \) <=> \(7x=21\) <=> \(x=21/7\) <=> \(x=3\)

2)\(\frac{ 9}{(- 3x)}=\frac{12}{(x + 10)}\) <=> \( (x+10)9 = 12(-3x)\)<=> \(9x + 90 = -36x\)<=>\(-45x=90 \) <=> \(x=-2\)

3) \(\frac{ (j + 4)}{j}+2 = 2 -\frac{1}{j}\),\(j≠ 0\)<=> \(\frac{(j+4)}{j} = -\frac{1}{j}\)<=> \(j+4=-1\) <=> \(j=-5\)

4) \(\frac{8}{q }+2=\frac{(q+4)}{(q-1)}\),\(q≠0,q≠ 1\)<=> \(\frac{8+2q}{q }=\frac{(q+4)}{(q-1)}\)<=>\((q-1)(8+2q)=(q+4)(q)\)<=>\(8q+2q^2-8-2q=q^2+4q\) <=> \(q^2+2q-8=0\)<=>\((q-2)(q+4)=0 \)<=> \(q=2,q=-4\)

5) Answer \(B\) Because \(x-5\) is in the denominator and NOT \(x+1\)

 

Hope it helps! 

Dimitristhym Oct 19, 2018
 #2
avatar+100571 
+1

Excellent,  Dimiristhym   !!!!

 

 

cool cool cool

CPhill  Oct 19, 2018
 #3
avatar+322 
+1

Thank you CPhill!

Dimitristhym  Oct 19, 2018

14 Online Users