Asked by britbrat
ind two consecutive even integers such that the smaller added to five times the larger is 58
Answers
Answered by
MathMate
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)
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)
Answered by
Bosnian
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
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
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