SOLUTION: The sum of two numbers is 27. One half of the first number plus one third of the second number is 11. Find the numbers.

Question 388997: The sum of two numbers is 27. One half of the first number plus one third of the second number is 11. Find the numbers.
Found 3 solutions by haileytucki, mananth, richard1234:
Answer by haileytucki(390)   (Show Source): You can put this solution on YOUR website!
a+b=27_(1)/(2)*a+(1)/(3)*b=11
Multiply (1)/(2) by a to get (a)/(2).
a+b=27_(a)/(2)+(1)/(3)*b=11
Multiply (1)/(3) by b to get (b)/(3).
a+b=27_(a)/(2)+(b)/(3)=11
Since b does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting b from both sides.
a=-b+27_(a)/(2)+(b)/(3)=11
Replace all occurrences of a with the solution found by solving the last equation for a. In this case, the value substituted is -b+27.
a=-b+27_(-b+27)/(2)+(b)/(3)=11
Remove the parentheses around the expression -b+27.
a=-b+27_(-b+27)/(2)+(b)/(3)=11
Divide each term in the numerator by the denominator.
a=-b+27_-(b)/(2)+(27)/(2)+(b)/(3)=11
Combine the numerators of all expressions that have common denominators.
a=-b+27_(-b+27)/(2)+(b)/(3)=11
Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of 6. The ((-b+27))/(2) expression needs to be multiplied by ((3))/((3)) to make the denominator 6. The (b)/(3) expression needs to be multiplied by ((2))/((2)) to make the denominator 6.
a=-b+27_(-b+27)/(2)*(3)/(3)+(b)/(3)*(2)/(2)=11
Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 6.
a=-b+27_((-b+27)(3))/(6)+(b)/(3)*(2)/(2)=11
Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 6.
a=-b+27_((-b+27)(3))/(6)+(b(2))/(6)=11
The numerators of expressions that have equal denominators can be combined. In this case, ((-b+27)(3))/(6) and ((b(2)))/(6) have the same denominator of 6, so the numerators can be combined.
a=-b+27_((-b+27)(3)+(b(2)))/(6)=11
Simplify the numerator of the expression.
a=-b+27_(-b+81)/(6)=11
Multiply each term in the equation by 6.
a=-b+27_(-b+81)/(6)*6=11*6
Simplify the left-hand side of the equation by canceling the common factors.
a=-b+27_-b+81=11*6
Multiply 11 by 6 to get 66.
a=-b+27_-b+81=66
Since 81 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 81 from both sides.
a=-b+27_-b=-81+66
Add 66 to -81 to get -15.
a=-b+27_-b=-15
Multiply each term in the equation by -1.
a=-b+27_-b*-1=-15*-1
Multiply -b by -1 to get b.
a=-b+27_b=-15*-1
Multiply -15 by -1 to get 15.
a=-b+27_b=15
Replace all occurrences of b with the solution found by solving the last equation for b. In this case, the value substituted is 15.
a=-(15)+27_b=15
Multiply -1 by the 15 inside the parentheses.
a=-15+27_b=15
Add 27 to -15 to get 12.
a=12_b=15
This is the solution to the system of equations.
a=12_b=15
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
The sum of two numbers is 27. One half of the first number plus one third of the second number is 11. Find the numbers.
...
let the numbers be x & y
...
x+y =27..............1
x/2 + y/3 = 11
LCD = 6
multiply by 6
3x+2y=66.............2
multiply (1) by -3
-3x-3y=-81
add this to (2)
-3x+3x-3y+2y=-81+66
-y=-15
y=15
...
plug y in (1)
x=12
...
m.ananth@hotmail.ca
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
Suppose the two numbers are x and y. Then, x+y = 27. From the second statement,

Multiply the previous equation by 6 to clear the fractions (it makes life much easier...)

. Rewrite 3x + 2y as x + 2x + 2y (so we can use substitution)

Since 2x + 2y = 54,

x = 12. Substituting into the very first equation, we get y = 15.
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