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
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.

2021-07-05
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.