Question

Four times a first number decreased by a second number is 10. The first number increased by four times the second number is 28. Find the numbers. a) x=6,y=−14

b) x=4,y=−6
c) x=2,y=6
d) x=10,y=−2

Answer 1

The two numbers can be found by solving a system of two equations: 4x - y = 10 and x + 4y = 28. Solving these equations, we find that the first number is 4 and the second number is 6, but there is a discrepancy as the options include c) x=2, y=6, which is likely a typo, and should read c) x=4, y=6.

Step-by-step explanation:

The problem presented can be boiled down to two equations using algebra, which we can solve simultaneously to find the values of the first and second numbers. Let's denote the first number as x and the second number as y. According to the problem:

1. 4 times the first number decreased by the second number is 10: 4x - y = 10
2. The first number increased by four times the second number is 28: x + 4y = 28

Solving these two equations simultaneously will give us the value of x and y. Here's the step-by-step solution:

1. From the second equation, express x in terms of y: x = 28 - 4y.
2. Substitute the expression for x from step 1 into the first equation: 4(28 - 4y) - y = 10.
3. Simplify and solve for y: 112 - 16y - y = 10, so -17y = -102, thus y = 6.
4. Substitute y back into the expression for x: x = 28 - 4(6), which gives x = 4.

Therefore, the first number (x) is 4 and the second number (y) is 6, which corresponds to option c) x=2, y=6, although there seems to be a typo in the options as the solution should be x=4, y=6.