# Thread: One way of solving systems of linear equations is by adding a multiple of one equatio

1. ## One way of solving systems of linear equations is by adding a multiple of one equatio

There are two factors that limit how much he can bake in a week: He only wants to work for 40 a week and he only has one oven. Suppose that it takes the baker 1 to prepare a pair of cakes or a gross of cookies (before they are placed in the oven). Since he only wants to work 40 a week, his output of pairs of cakes and his output of grosses of cookies are constrained by the equation .
To maximize the profit of the bakery, the first step is to find where the equations for all of the constraints intersect. For the following part, you will look at and , which is also a constraint (specifically a minimum) since the baker cannot make a negative number of cookies.

What should you multiply the equation by so that when added to the variable will cancel out?

2. if y = 0 , and x+y = 40, then x = 40. no "multiplication" is necessary.

seems your problem statement for profit maximization is missing some other constraint information.

3. Suppose that a baker makes cakes and cookies. He knows that he is most efficient when he makes pairs of cakes (instead of one cake at a time) or a dozen dozens of cookies. Call the number of pairs of cakes that he bakes in a week x and the number of grosses (dozens of dozens) of cookies that he bakes in a week y.

Here's the beginning part to the question that I forgot.