Math ProjectMath Project

Josh Gustavus Josh Gustavus Candyce CooleyCandyce CooleyMichael Lipsey Michael Lipsey

Craig WrightCraig Wright

The Problem…The Problem…

►A militia preparing for a raid needs 2328 A militia preparing for a raid needs 2328 lbs. of gun powder with an ignition rate of lbs. of gun powder with an ignition rate of 40%. 40%.

►Before they could carry out their operation, Before they could carry out their operation, their base was raided and 1028 lbs. of their base was raided and 1028 lbs. of their pre mixed gunpowder was taken. their pre mixed gunpowder was taken.

►The militia was left with 1300 lbs. of gun The militia was left with 1300 lbs. of gun powder with an ignition rate of 40% and powder with an ignition rate of 40% and they also had gun powder with an ignition they also had gun powder with an ignition rate of 20%.rate of 20%.

The Solution…The Solution…

► In order to get the In order to get the 20% back to the 20% back to the 40%, they need, 40%, they need, they had to re-they had to re-supply and use supply and use some of the 70% some of the 70% ignition rate from ignition rate from their home base to their home base to mix. mix.

The Formula…The Formula…

► The militia has 1300 lbs. of gun powder left The militia has 1300 lbs. of gun powder left and needs a total of 2328 lbs. so they need to and needs a total of 2328 lbs. so they need to mix 1028 lbs.mix 1028 lbs.

► To find out how many pounds of each percent To find out how many pounds of each percent they need you use the equation X+Y=1028they need you use the equation X+Y=1028

► Sense the desired rate of the mixture needs Sense the desired rate of the mixture needs to be 40% and the two available batches of to be 40% and the two available batches of gunpowder have ignition rates of 70% and gunpowder have ignition rates of 70% and 20% the second equation will 20% the second equation will be .70x+.20y=1028(.40) be .70x+.20y=1028(.40)

How to solve…How to solve…

►For these types of problems there are For these types of problems there are two ways to solve them you can use two ways to solve them you can use either the Substitution Method or the either the Substitution Method or the Elimination Method.Elimination Method.

►For this project we will show both For this project we will show both methodsmethods

Elimination Method…Elimination Method…

► First we multiply our second equation by 10 to get First we multiply our second equation by 10 to get rid of decimals leaving us with 7x+2y=4112rid of decimals leaving us with 7x+2y=4112

► Next we multiply our first equation by -2 and Next we multiply our first equation by -2 and subtract the solution from are second problem. subtract the solution from are second problem.

► -2(x+y=1028) this gives us -2(x+y=1028) this gives us

(-2x-2y= -2056) (-2x-2y= -2056) -(7x+2y= 4112)-(7x+2y= 4112)

-5x=-2056-5x=-2056

► If we then divide both sides by -5 we can figure out If we then divide both sides by -5 we can figure out that x=411.2 that x=411.2

Elimination Method…Elimination Method…

► If we take the 411.2 and plug it in for x If we take the 411.2 and plug it in for x in the equation x+y=1028 we can in the equation x+y=1028 we can then solve for y.then solve for y.

►411.2+y=1028 411.2+y=1028 ►Subract 411.2 from both sides and we Subract 411.2 from both sides and we

have y=616.8have y=616.8

Substitution method…Substitution method…

► For this method we take the same two For this method we take the same two formulas x+y=1028 and 7x+2y=4112formulas x+y=1028 and 7x+2y=4112

► First we need to isolate a variable in the first First we need to isolate a variable in the first equation by subtracting y from both sides to equation by subtracting y from both sides to get x=1028-y get x=1028-y

►Next we substitute that solution into the Next we substitute that solution into the second equation giving us second equation giving us 7(1028-y)+2y=41127(1028-y)+2y=4112

►Distribute the seven 7196-7y+2y=4112 Distribute the seven 7196-7y+2y=4112 then combine like terms to get 7196-then combine like terms to get 7196-5y=41125y=4112

Substitution method…Substitution method…

►Next we subtract 7196 from both sides Next we subtract 7196 from both sides leaving us with -5y=-3084leaving us with -5y=-3084

► Finally we dived both sides by -5 and learn Finally we dived both sides by -5 and learn that y=616.8that y=616.8

►Now we can take that information and plug Now we can take that information and plug it in to the first problem and have it in to the first problem and have x+616.8=1028 x+616.8=1028

► So x=411.2So x=411.2

Conclusion…Conclusion…

►So in order for the militia to get its So in order for the militia to get its gunpowder supply back to the full gunpowder supply back to the full 2328 pounds needed, with an ignition 2328 pounds needed, with an ignition rate of 40%, they must mix together rate of 40%, they must mix together 411.2 pounds of powder with an 411.2 pounds of powder with an ignition rate of 70% and 616.8 pounds ignition rate of 70% and 616.8 pounds of powder with an ignition rate of 20%of powder with an ignition rate of 20%

Class Problem…Class Problem…

► Lets say instead of needing 2328 lbs of Lets say instead of needing 2328 lbs of gunpowder with an ignition rate of 40% they gunpowder with an ignition rate of 40% they need 1750 lbs. of gun powder with an ignition need 1750 lbs. of gun powder with an ignition rate of 70%. rate of 70%.

► 735 lbs of their powder was taken.735 lbs of their powder was taken.► In their storage they have powder with an In their storage they have powder with an

ignition rate of 80% and an ignition rate of 30% ignition rate of 80% and an ignition rate of 30% ►How much of each type of gun powder do they How much of each type of gun powder do they

need to mix to be back at 1750 lbs. of gun need to mix to be back at 1750 lbs. of gun powder with an ignition rate of 70%powder with an ignition rate of 70%

The Answer…The Answer…

►203 lbs of gun powder with an ignition 203 lbs of gun powder with an ignition rate of 80%rate of 80%

►812 lbs. of gun powder with an 812 lbs. of gun powder with an ignition rate of 30% ignition rate of 30%