x = first number, y = second number
sum is 62, means x+y=62
difference is 10, means y-x = 10
You can solve a system of equations by Substitution, Elimination, Graphing, or Matrices.
Graphing is probably not the best for this one, but it is possible by putting both equations in y=mx+b form and putting into a graphing calculator.
y=-x+62
y=x+10
Substitution:
x+y=62
x+y-y = 62-y
x=62-y
Now substitute for x in the second equation
y-x=10
y-(62-y) = 10
distribute the negative
y-62+y=10
combine like terms
2y-62=10
2y-62+62 = 10+62
2y = 10+62
2y = 72
(2y)/2 = 72/2
y=36
Now substitute y into original equation (either will work)
y-x=10
36-x=10
-x=10-36
-x=-26
x=26
Elimination:
x+y=62
y-x=10
x +y = 62
-x +y = 10 Now add the two equations together
0 2y = 72
2y/2 = 72/2
y=36
substitute y into either equation as in substitution (either works)
x+y=62
x+36=62
x=62-36
x=26
Matrices:
Put the Coefficients and constants into a matrix as follows:
x +y = 62
-x +y = 10
[ 1 1 | 62 ]
[-1 1 | 10]
either multiply a row by a number or add a row by the other row multiplied by a number.
Get the matrix in the form
[1 0 | x]
[0 1 | y]
Easier to start with the a 0, so get the -1 in the second row to 0
[ 1 1 | 62 ]
[-1+(1*1) 1+(1*1) | 10+(1*62) ]
[ 1 1 | 62 ]
[ 0 2 | 72 ]
[ 1 1 62 ]
[ 0*(1/2) 2*1/2) | 72*(1/2)]
[ 1 1 | 62 ]
[ 0 1 | 36 ] <-- we now have the y value
[ 1+(-1*0) 1+(-1*1) | 62+(-1*36) ]
[ 0 1 | 36 ]
[ 1 0 | 26 ] <-- we now have x value
[ 0 1 | 36 ] <-- and we have y value
x=26, y=36