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(6 pts) Bart has money invested in Iwo funds: stock fund and bond fund. Last year the stock find paid 8"_ the bond fund paid 2% and he received S780 This year ...

Question

(6 pts) Bart has money invested in Iwo funds: stock fund and bond fund. Last year the stock find paid 8"_ the bond fund paid 2% and he received S780 This year the stock fund paid 10%. the bond fund paid 1% and he received S810. How much money does he have in cach fund? You must define two variables (write them down: Let T amount at 8%" . etc_ set up syslem of two linear equations in these variables. solve the system by hand, and write the solution in complete sentence(6 pts) A minor le

(6 pts) Bart has money invested in Iwo funds: stock fund and bond fund. Last year the stock find paid 8"_ the bond fund paid 2% and he received S780 This year the stock fund paid 10%. the bond fund paid 1% and he received S810. How much money does he have in cach fund? You must define two variables (write them down: Let T amount at 8%" . etc_ set up syslem of two linear equations in these variables. solve the system by hand, and write the solution in complete sentence (6 pts) A minor league baseball stadium has 7000 seats. Box seats cost S6, grandstand seats cost $4 and bleacher seats cost S2 When all seats are sold, the revenue is S26.400. The number ofbox seats is one-third the number of bleacher seats_ How many seats of each type are there? You must define three variables (write them down: "Let x = box seats' etc ) , set up system of three linear equations in these variables. solve the syslem by hand. and wrile the solution in complete sentence Note: Do not define your variables by saying "x =6,y=4,2-27 (12 pts) Given the following matrices: A = [5 8=[ & cl; perform each of these matrix operations or stale why it can be done: 2A 3C = AB = (8 pts) Use the method of matrix inverses t0 solve the following system. You must invert the coefficient matrix by hand and then solve the system to get all of the points_ 3x + 2y 62 = 3 x+y+22 = 2 2x + 2y + S2 = 0



Answers

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to $1 .$ It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction? (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (iii) The sum of the digits of a two-digit number is 9 . Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (iv) Meena went to a bank to withdraw $? 2000$. She asked the cashier to give her $? 50$ and $? 100$ notes only. Meena got 25 notes in all. Find how many notes of $? 50$ and $? 100$ she received. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid $? 27$ for a book kept for seven days, while Susy paid $? 21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

This problem is asking us to solve a system of equations by using the inverse of a coefficient matrix. Now, this problem is very similar to a previous problem. And so all of the work I have shown shown right here is from problems 61. The only difference in this problem is that this value is different. The 705 is now changed to 760. And so already having done a lot of work on the previous problem, what we can do is simply erase the 705 and right in 760. This is the only difference between this problem and the previous all the hard work that we did calculating the inverse matrix or in my case, just plugging it into a calculator has already been done. And we don't need to do any more work with that. Now, all we have to do is multiply through the Matrix with their new value 760. And so, instead of the previous values that we got well and get these ones when I got is 4000, 2000 and 4000 again, again, if you want to see how I got the inverse matrix or how I changed the system of equations to a system of matrices. Just check out the previous problem, and so this is our final answer.

This problem is asking us to solve a system of equations using the inverse of a coefficient matrix. It's also very similar to another problem in this chapter, namely problem 61. But I currently have on the screen is work that I did solving problems 61 the only difference here is to values. So if you wanted to see how I got the English Matrix or how I solved how I turned the system of equations, two matrices go back to problem 61 but already haven't done that work. All we have to do is erase these two values right here. The 10,000 and the 705 from Problems 61 are now different values. So I'll erase those here, and what we have instead, for this problem is 12,000 here and 835 year. And so I'll just go down to my matrix speed and I'll write down the same values and I'll copy them down here. So now, with our new values, we can multiply a inverse with our new values and be and we can solve our system of equations and we get 9000 for ex 1000 for why? And lastly, 2000 for a Z. And this is our final answer

Mhm. This problem is asking us to solve a system of equations using the inverse of the coefficient matrix. It's also very similar to Problems 61 in the same section, so the existing work on the screen is coming from that problem. So if you wanted to see how I changed the system of equations to an equations using matrices, or if you wanted to see how I got a inverse, just go back to problems 61. The only difference between that problem and this one is that we have two new values. First, the total investment is now 500,000, and so I'll write 500,000 and all of the spots that it should be. Next. We have a new annual return, and that is 38,000. And so, like, why? It's alright. That and all of this phone stuff this value should be. And there we go. Now we have our new Matrix B and so we can do a inverse times B, and we will get the values that we need to solve the system of equations. So first we have X is equal to 200,000. We'll get why is equal to 100,000. And lastly, Z is equal to 200,000 as well. And this right here is our final answer.

In this question, we're solving a system of equations using the inverse of the coefficient matrix. And so what that means is that we have some kind of problem. Where we have a X is equal to B or A and B are matrices and X are the very X is the variables. What we want to do is we want to solve for X, and that would be equal to a inverse times B. And so the equations are the system of equations that were given is as follows we have X plus y plus Z is equal to 10,000. We also have 0.65 x plus 0.7 y plus 0.9 z is equal to 705. And lastly, we have till why minus Z is equal to zero, and what we can do here is we can add a zero X, and this way it'll make more sense when we turn it into a matrix. And so when we're turning into a matrix, we're just going to extract all of the constant values so we'll have ones all across the top, and then we'll have all of the decimal coefficients here. And then this is where the zero comes in. We'll have zero to a negative one in our last row. Next comes all of our variables, and we'll just do X, Y and Z like this, and we'll set this equal to what was equal above all of our answers are all those constant values at the end of those equations above. So we'll have 10,000, 705 and zero, and this is the same set up as before. Up here we have our matrix, a all over variables X, and this is a B. And so what we want to do is we want to find a inverse now. What you could do is you could set up a an augmented matrix here where you have a on one side, the identity matrix on the other, and you do a lot of row operations to get the identity matrix on the left and then you'll you'll end up with a inverse on the right. Now that's one method you could do because it's a three by three matrix. It will take a little while. And so what I did is I just found a matrix calculator. And that is how I found my values for a inverse. And so I'll write it down here. So what I have for a inverse is I got all of these values. I got 500 over 11 negative, 600 over 11 negative, 4/11, like 13/11, 200 over 11, 5/11. It's a five. There we go. Negative 26/11, 400 over 11 and lastly, negative 1 11. And so that is our A inverse. And following this equation up here above, let's see if I can circulate. There we go. Um, I will multiply a inverse by here, So let me erase this part. It's not only any sense, so I'll have a inverse Times B is equal to this. First Matrix is a inverse. Then we'll right be just as it is down below. And to do this operation, you would multiply these values up here by the these three right here. And so the 10,000 will be multiplied with the 61st is 705 will be multiplied by the second value, and then you'll have zero times the last value. And then you repeat that for the other two rows, and then you will get your solution for X. What I got is 7000, 1000 and 2000, and this is your solution for the system of equations.


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