Question

One number is eleven more than another. If their sum is increased by seventeen, the result is$90$. Find the numbers.

Short answer

One number is eleven more than another and their sum is increased by seventeen, the result is$90$then the numbers are$31\&42.$

Step-by-step solution

Step 1. Given information   

The second number is eleven more than the first number.

Step 2. Consider variable for numbers

Consider the first number is n

The second number is eleven more than the first number.

Then the second number is $n+11$

Step 3. Solution 

The sum of the first number, second number and $17$is $90$

$$\begin{gathered} n+(n+11)+17=90 \\ 2n+28=90 \\ 2n=62 \\ n=31 \end{gathered}$$

The first number is $n=31$

the second number is

$$\begin{gathered} n+11=31+11 \\ n+11=42 \end{gathered}$$

So the numbers are$31\&42$

Step 4. Verification

$42$is $11$more than$31$.

$$\begin{gathered} 42\stackrel{?}{=}31+11 \\ 42=42 \end{gathered}$$

The left-hand side is equal to the right-hand side

The sum of $31,42,\&17$is $90$

$$\begin{gathered} 31+42+17\stackrel{?}{=}90 \\ 90=90 \end{gathered}$$

The left-hand side is equal to the right-hand side

so our answer is correct

the numbers are$31\&42$