The sum of two numbers is $56$56 and their difference is $30$30.
Set up two equations by letting $x$x and $y$y be the two numbers.
Use $x$x as the larger of the 2 numbers.
Sum equation: | $\editable{}$ |
Difference equation: | $\editable{}$ |
First solve for $x$x.
Equation 1 | $x+y=56$x+y=56 |
Equation 2 | $x-y=30$x−y=30 |
Now solve for $y$y.
We want to find the value of two numbers, $x$x and $y$y. We know that:
The length of a rectangle measures $12$12 units more than the width, and the perimeter of the rectangle is $56$56 units.
Let $y$y be the width and $x$x be the length of the rectangle.
There are $36$36 members in a group, and the men outnumber the women by $16$16.