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What is the zero of
q(x) = -7/3x + 14?

User Nulik
by
5.3k points

1 Answer

3 votes

Answer:

x = 6

Explanation:

Solve for x:

-(7 x)/3 + 14 = 0

Put each term in -(7 x)/3 + 14 over the common denominator 3: -(7 x)/3 + 14 = -(7 x)/3 + 42/3:

(-(7 x)/3 + 42/3) = 0

-(7 x)/3 + 42/3 = (-7 x + 42)/3:

((-7 x + 42)/3) = 0

Multiply both sides of (-7 x + 42)/3 = 0 by 3:

(3 (-7 x + 42))/3 = 3×0

(3 (-7 x + 42))/3 = 3/3×(-7 x + 42) = -7 x + 42:

(-7 x + 42) = 3×0

0×3 = 0:

-7 x + 42 = 0

Subtract 42 from both sides:

(42 - 42) - 7 x = -42

42 - 42 = 0:

-7 x = -42

Divide both sides of -7 x = -42 by -7:

(-7 x)/(-7) = (-42)/(-7)

(-7)/(-7) = 1:

x = (-42)/(-7)

The gcd of 42 and -7 is 7, so (-42)/(-7) = (-(7×6))/(7 (-1)) = 7/7×(-6)/(-1) = (-6)/(-1):

x = (-6)/(-1)

(-6)/(-1) = (-1)/(-1)×6 = 6:

Answer: x = 6

User Zahid M
by
4.7k points