## Algebra 1: Common Core (15th Edition)

This problem is a simple system of equations. This can be set up as: Legend: x=19 and y=8, in terms of the values on the dartboard. 1st: $x+y=7$ 2nd: $19x+8y=100$. Using multiplication to achieve the same coefficients for y, we can multiply the 1st equation by 8. This will give us an equation of $8x+8y= 56$. We can then subtract the 2nd equation from this new equation. $8x+8y= 56$ - $19x+8y=100$ This will result in $-11x= 44$. Dividing both sides by -11 will give a value of $x=4$. Substituting x=4 into the first equation will allow us to solve for y. $4+y=7$ which simplifies to $y=3$ This solutions means that 100 can be scored with 4 19's and 3 8's.