Find two mixed numbers so that the sum is 15 3/10 and the difference is 8 5/10
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What is this
Bobpursley just doesn't know the answer to b
To find two mixed numbers that satisfy the given conditions, let's start by assigning variables to represent the two numbers.
Let's call the first mixed number x and the second mixed number y.
According to the given conditions, we have two equations:
Equation 1: x + y = 15 3/10
Equation 2: x - y = 8 5/10
To solve this system of equations, we can use the method of substitution or elimination.
Let's start by using the substitution method. We can rearrange Equation 2 to get x = y + 8 5/10.
Now, substitute this expression for x in Equation 1:
(y + 8 5/10) + y = 15 3/10
Simplify the equation by adding the fractions separately from the whole numbers:
2y + 8 5/10 = 15 3/10
To eliminate the fractions, convert the mixed numbers to improper fractions:
2y + 8 1/2 = 15 1/3
Converting the mixed numbers to fractions:
2y + 17/2 = 46/3
To clear the fractions, multiply each term by the least common denominator (2*3=6):
6(2y + 17/2) = 6(46/3)
12y + 51 = 92
Now, isolate the variable y by moving the constant term to the other side of the equation:
12y = 92 - 51
12y = 41
Finally, divide by 12 to solve for y:
y = 41/12
To find the value of x, substitute this value of y back into Equation 2:
x - 41/12 = 8 5/10
Convert the whole number to a fraction with the same denominator as 12:
x - 41/12 = 8 1/2
Convert the mixed number to an improper fraction:
x - 41/12 = 17/2
To combine the whole numbers and fractions, convert 8 to an improper fraction:
x - 41/12 = 34/4
Simplify the equation by multiplying all terms by 12 to eliminate the fractions:
12x - 41 = (34/4) * 12
12x - 41 = 102
Add 41 to both sides of the equation to isolate x:
12x = 102 + 41
12x = 143
Divide by 12 to solve for x:
x = 143/12
Therefore, the two mixed numbers that satisfy the given conditions are x = 143/12 and y = 41/12.
A-B=8.5
A+B=10.333333
add the equations.
2A=18 8/10
A=9 4/10
now figure out B yourself