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# rational equations

0
376
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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

#1
+2

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
+2

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
+1

Excellent,  Dimiristhym   !!!!   CPhill  Oct 19, 2018
#3
0

Thank you CPhill!

Dimitristhym  Oct 19, 2018