+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)\(\frac{ (x + 7)}{2 }=\frac{5x}{3}\) <=> \(3 (x + 7) = 2(5x)\) <=> \(3x + 21 = 10x \) <=> \(7x=21\) <=> \(x=21/7\) <=> \(x=3\)
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!
+343 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!
