ind two consecutive even integers such that the smaller added to five times the larger is 58
a = first number
b = second number
b = a + 2
a + 5 ( a + 2 ) = 58
a + 5 a + 10 = 58
6 a + 10 = 58 Subtrac 10 to both sides
6 a + 10 - 10 = 58 - 10
6 a = 48 Divide both sides by 6
6 a / 6 = 48 / 6
a = 8
b = a + 2 = 8 + 2 = 10
Your numbers are 8 and 10
8 + 5 * 10 = 8 + 50 = 58
Let the smaller even integer be n.
The larger is then n+2.
The problem says
"smaller added to five times the larger is 58"
which translates to
n+5(n+2)=58
Solve for n (smaller even integer)
Let's represent the smaller even integer as x, and the larger even integer as x + 2 (since they are consecutive even integers).
According to the given information, we can form the equation:
x + 5(x + 2) = 58
Now, let's solve this equation step-by-step:
Step 1: Distribute the 5 to both terms inside the parentheses.
x + 5x + 10 = 58
Step 2: Combine like terms.
6x + 10 = 58
Step 3: Subtract 10 from both sides to isolate the term with variable.
6x = 58 - 10
6x = 48
Step 4: Divide both sides by 6 to solve for x.
x = 48 / 6
x = 8
So, the smaller even integer is 8.
Step 5: Find the next consecutive even integer by adding 2 to the smaller even integer.
x + 2 = 8 + 2
x + 2 = 10
So, the larger even integer is 10.
Therefore, the two consecutive even integers that satisfy the given conditions are 8 and 10.
To find two consecutive even integers, let's call the smaller one "x" and the larger one "x+2."
According to the problem, the smaller integer added to five times the larger integer is 58.
So we can write the equation as follows:
x + 5(x+2) = 58
Now, let's solve the equation step by step:
x + 5x + 10 = 58
6x + 10 = 58
Subtract 10 from both sides:
6x = 48
Divide both sides by 6:
x = 8
Now, we know that the smaller even integer is 8.
To find the larger even integer, we can substitute the value of x back into the equation:
x + 2 = 8 + 2 = 10
Therefore, the two consecutive even integers that satisfy the given condition are 8 and 10.