Question

The sum of three times a first number and twice a second number is 43. If the second number is subtracted from twice the first number, the result is -

Upper level algebra
ANSWERED
asked 2021-07-03
The sum of three times a first number and twice a second number is 43. If the second number is subtracted from twice the first number, the result is -4. Find the numbers.

Expert Answers (2)

2021-07-05
36
 
Best answer
2021-08-04

Let x be the first number and yy be the second number.
The sum of three times a first number and twice a second number is 43 so:
\(\displaystyle{3}{x}+{2}{y}={43}{\left({1}\right)}\)
If the second number is subtracted from twice the first number, the result is -4 so:
\(\displaystyle{2}{x}−{y}=−{4}{\left({2}\right)}\)
Solve by substitution. Solve for yy using (2) to obtain (3):
\(\displaystyle{y}={2}{x}+{4}{\left({3}\right)}\)
Substitute (3) to (1) and solve for xx:
\(\displaystyle{3}{x}+{2}{\left({2}{x}+{4}\right)}={43}\)
\(\displaystyle{3}{x}+{4}{x}+{8}={43}\)
\(\displaystyle{7}{x}+{8}={43}\)
\(\displaystyle{7}{x}={35}\)
\(\displaystyle{x}={5}\)
Solve for yy using (3):
\(\displaystyle{y}={2}{\left({5}\right)}+{4}\)
\(\displaystyle{y}={14}\)
So, the numbers are 5 and 14.

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