seriously needing someone to help me with some confusing math!
OK, it is finally time to solve our problem about the weights of beluga whales and orca whales. Question: The weights of two beluga whales and three orca whales totals 36,000 pounds. The weights of one of the belugas and one of the orcas add up to 13,000 pounds. Using this information only, how much do the belugas weigh? How much do the orcas weigh? Part 1: Write a system of equations for this problem and solve it by graphing. (You can write 36,000 and 13,000 as 36 and 13 respectively as long as you remember to put the thousand back on in your answer.) You can download a graphing program (information is provided in the course information area) to create your graph or just draw them on paper. Be sure to indicate the point of intersection on the graph by writing your answer as an ordered pair. Part 2: Using the same system of equations, solve it by substitution. Be sure to show all work in this process; a final answer is not enough to earn credit. Answer the question in complete sentences. Part 3: Solve the same system of equations by the addition method. Again, be sure to show all work and indicate your answers in complete sentences.
Let the weight of a beluga whale be b and of an orca be a (measured in thousand pounds). 2b + 3a = 36 ...(1) b + a = 13 ...(2) From (2): b = 13 - a ...(3) Substituting for b in (1) from (3) in (1): 2(13 - a) + 3a = 36 26 - 2a + 3a = 36 3a - 2a = 36 - 26 a = 10 The orcas weigh 10,000 lb. Substituting for a in (3): b = 13 - 10 = 3. The bellugas weigh 3,000 lb.
okay
so
Lol, helpful?
haha not shure it aint registering lol
so for part one my equation would be 2b + 3a = 36 b + a = 13
Yes
@rational
Okay, so what I gave you was for Part Two.
Part One: 2b + 3a = 36 ...(1) b + a = 13 ...(2)
Part Three: 2(13 - a) + 3a = 36 26 - 2a + 3a = 36 3a - 2a = 36 - 26 a = 10 Use this to plug in to your (a) & (b). The use addition to get the final weights of Orca's to Beluga's
*Then use...
okay so I for every A i substitute it with 10 and there isn't and B's so i dont have to worry bout those right?
@confluxepic can you help me finish up?
@Data_LG2 @JFraser @butterflydreamer
Part 1: You already done with the equation, but you still need to graph those equations. Part 2: is the substitution method, which riley already provided. Part 3: Refers to a different method, which they called addition method (I assume it's the elimination method). You have to solve the system in a different way
okay so how do i do it with the addition method
2b + 3a = 36 -> eqn (1) b + a = 13 -> eqn (2) 1. Multiply the 2nd equation by -2 (for each term multiply them by -2) 2. add equation 1 and eqn 2 (variable b should be eliminated and you should get the value of a) 3. Substitute the value of a that you got from step 2 in any of eqn 1 or eqn2. 4. Simplify and solve for b.
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