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What is 5x+538/23*1000-500=?

1 Answer

5 votes

Answer:

x = (-105300)/23

Explanation:

Solve for x:

5 x + 526500/23 = 0

Hint: | Put the fractions in 5 x + 526500/23 over a common denominator.

Put each term in 5 x + 526500/23 over the common denominator 23: 5 x + 526500/23 = (115 x)/23 + 526500/23:

(115 x)/23 + 526500/23 = 0

Hint: | Combine (115 x)/23 + 526500/23 into a single fraction.

(115 x)/23 + 526500/23 = (115 x + 526500)/23:

(115 x + 526500)/23 = 0

Hint: | Multiply both sides by a constant to simplify the equation.

Multiply both sides of (115 x + 526500)/23 = 0 by 23:

(23 (115 x + 526500))/23 = 23×0

Hint: | Cancel common terms in the numerator and denominator of (23 (115 x + 526500))/23.

(23 (115 x + 526500))/23 = 23/23×(115 x + 526500) = 115 x + 526500:

115 x + 526500 = 23×0

Hint: | Any number times zero is zero.

0×23 = 0:

115 x + 526500 = 0

Hint: | Isolate terms with x to the left hand side.

Subtract 526500 from both sides:

115 x + (526500 - 526500) = -526500

Hint: | Look for the difference of two identical terms.

526500 - 526500 = 0:

115 x = -526500

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 115 x = -526500 by 115:

(115 x)/115 = (-526500)/115

Hint: | Any nonzero number divided by itself is one.

115/115 = 1:

x = (-526500)/115

Hint: | In (-526500)/115, the numbers 526500 in the numerator and 115 in the denominator have gcd greater than one.

The gcd of 526500 and 115 is 5, so (-526500)/115 = (-(5×105300))/(5×23) = 5/5×(-105300)/23 = (-105300)/23:

Answer: x = (-105300)/23

User Nikhil Swami
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