+642 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.
+343 1)(x+7)2=5x3 <=> 3(x+7)=2(5x) <=> 3x+21=10x <=> 7x=21 <=> x=21/7 <=> x=3
2)9(−3x)=12(x+10) <=> (x+10)9=12(−3x)<=> 9x+90=−36x<=>−45x=90 <=> x=−2
3) (j+4)j+2=2−1j,j≠0<=> (j+4)j=−1j<=> j+4=−1 <=> j=−5
4) 8q+2=(q+4)(q−1),q≠0,q≠1<=> 8+2qq=(q+4)(q−1)<=>(q−1)(8+2q)=(q+4)(q)<=>8q+2q2−8−2q=q2+4q <=> q2+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!
+343 1)(x+7)2=5x3 <=> 3(x+7)=2(5x) <=> 3x+21=10x <=> 7x=21 <=> x=21/7 <=> x=3
2)9(−3x)=12(x+10) <=> (x+10)9=12(−3x)<=> 9x+90=−36x<=>−45x=90 <=> x=−2
3) (j+4)j+2=2−1j,j≠0<=> (j+4)j=−1j<=> j+4=−1 <=> j=−5
4) 8q+2=(q+4)(q−1),q≠0,q≠1<=> 8+2qq=(q+4)(q−1)<=>(q−1)(8+2q)=(q+4)(q)<=>8q+2q2−8−2q=q2+4q <=> q2+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!
