A university bookstore recently sold a wirebound graph-paper notebook for $0.88, and a college-ruled notebook for $2.38. At the start of spring semester, a combination of 50 of these notebooks were sold for a total of $59.00. How many of each type were sold?
What are your givens?
List them out first.
graph paper=40, college ruled=10
g=graph paper, c=college-ruled. $0.88g+$2.38c=$59.00 (a total of $59.00 was made) g+c=50 (a total of 50 was sold) use substitution method. c=50-g 0.88g+2.38(50-5)=59 0.88g+119-2.38g=59 -1.5g+119=59 -1.5g=-60 (-1.5/-1,5)g=-60/-15 g=40 40+c=50 -40 -40 c=10
Substitution is unnecessary and a complete waste of time in this situation. Just eliminate.
can you show me? petewe
and thank you nahomie
Let x = graph let y = college 0.88x + 2.38y = 59 x + y = 50 What do you from there if you want to eliminate a variable?
add/subtract that variable so they canel
Yes but because the variables in each equation have different coefficients, how do we make it so that at least 1 variable in each equation is equal?
that is what has me confused..i dont know
If you change one side of an equation, what must be done to the other?
you have to change that side too
Exactly. Given that, how can we make the x in the second equation equal to the x in the first equation? 0.88x + 2.38y = 59 x + y = 50
add them
No. What do you have to do x to make it 0.88x?
multiply???
Good, now if you multiply x, what do you have to do to everything else in the equation?
multiply as well
Exactly, show me what you get.
0.88x+2.09=51.92
Good, now the x in both equations are equal. From here, subtract the equations and the x will cancel out. Isolate and solve for y. Once you have y, sub that into x + y = 50 to find x. Hope I helped.
you did THANK YOU!
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