Question

ricardo has two types of assignments for his class. the number of mini assignments, m, he has is 1 fewer than twice the number of long assignments, l, he has. if he has 46 assignments in total, which of the following systems of equations can be used to correctly solve for m and l? choose 1 answer: (a) m = 2l - 1 m + l = 46 (b) m = 2l - 1 m = l + 46 (c) l = 2m - 1 m + l = 46 (d) l = 2m - 1 m = l + 46

Explanation:

Step1: Translate the first - condition

The number of mini - assignments $m$ is 1 fewer than twice the number of long assignments $l$. So, $m = 2l-1$.

Step2: Translate the second - condition

The total number of assignments is 46. So, the sum of mini - assignments $m$ and long assignments $l$ is 46, which gives $m + l=46$.

Answer:

A. $m = 2l - 1$, $m + l = 46$