one number is 10 more than another. if the sum of twice the smaller number and 3 times the larger number is 55, find the two numbers.
2007-08-09 20:23:56 · 5 answers · asked by moni 1 in Education & Reference ➔ Homework Help
a- larger number
b- smaller number
a = b+10
2b+3a = 55
substitute
2b+3(b+10) = 55
distribute
2b+3b+30 = 55
Simplify
5b+30 = 55
Subtract 30 from both sides
5b = 25
b = 5
Substitute back into the first equation
a = b+10
a = 5+10
a= 15
small number = 5
larger number = 15
2007-08-09 20:34:50 · answer #1 · answered by Leila 2 · 0⤊ 0⤋
Let the 2 numbers be x and y
y= x+10
2x+3y =
2x+3(x+10) =
2x+3x+30 =
5x+30 =55
5x= 55-30=25
x= 25/5 =5
so y= x+10= 5+10=15
2007-08-10 06:59:59 · answer #2 · answered by varshanum1 2 · 0⤊ 0⤋
Smaller - x
Larger - (x+10)
2(x) + 3(x + 10) = 55
2x + 3x + 30 = 55
5x = 25
x = 5
2007-08-10 03:31:04 · answer #3 · answered by Kenshin 4 · 0⤊ 0⤋
Let the two numbers be X and Y
The first sentence tells you that
Y = X + 10
The second sentence tells you that:
2X+3Y = 55
You now have two equations in two unknowns. If you know linear algebram it is easy to solve.
If you don't -- then replace Y with (X+10) in the second equation. This gives you:
2X+3(X+10) = 55
or
2X+3X+30 = 55
or 5X + 30 = 55
You should be able to do the rest yourself.
2007-08-10 03:30:38 · answer #4 · answered by Ranto 7 · 0⤊ 0⤋
x= first number
x + 10 = second number
2x + 3(x + 10) = 55
5x + 30 = 55
5x = 25
x =5
x + 10 = 15
2007-08-10 03:32:48 · answer #5 · answered by Wise@ss 4 · 0⤊ 0⤋