Guide :

# find a general solution and check your answer by substitution 10y"+6y'-4y=0

solve bu subsitution method

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### Problem Set 2: Answer Key Econ 203 Intermediate ...

... Answer Key Econ 203 Intermediate Microeconomics Please show all the work ... both if we want to check our answer: ... MC = TC ′ = y3 − 2y 2 + 10y = 3y 2 ...

no solution. 5. 4x. 4. y. 1 4y. 12 9 (3, 0) 3x. 2 3x. 5x 2x. 9y 3y (4 ... Round your answer to the nearest tenth. 5.0. 1021 ergs. ... Check your solution by any ...

### Geometry and Discrete Mathematics - Scribd

Using Geometry and Discrete Mathematics ... Solution The circumference of the ... Find the coefficient of x10 in the binomial expansion of 2x simplify your answer. 5.

1. -2x2 + y2 + 4x + 6y + 3 = 0 3. x2 + 16y2 -64y = 0 5. x2 + y2 = -4x + 6y ... 20 = 0 b. x2 + y2 - 4y = 0 c. x2 + y2 - 6x - 10y = 2. ... Justify your answer. f ...

### Math 100 | Fraction (Mathematics) | Abstract Algebra

+ 6y −9 = 2y(4y 2 + 3) −3(4y 2 ... 0 , y 0 ), is a solution of the above system if each ... expressing your answer as

### Free Math Worksheets, Word Problems and Teaching Resources

... "><strong><span style="color: red; font-family: &quot;Trebuchet MS&quot;, sans-serif;">Solution: ... 0, can be find using ... and check your answer using ...

### 7 - Puzzle Museum

Answer is: 0 + 1 + 1 + 2 + 3 ... mathematicians were unable to get the correct solution and failed to check the solutions that ... general solution of: ...

### Visual Presentation in Algebra for First Year Students ...

Developing Skills in Algebra for First Year Students. Login ... 4.2y-x+4=0 -2x= -4y-8 ... show all work and show the calculations that check or test your solution ...

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## Suggested Questions And Answer :

### 6x+y−z=−3 ; 29x+5y+3z=6 ; 35x+6y+2z=3

Call the three equations A, B and C. Note that if we double C we get the same constant 6 as B. So we can write B=2C. That is, 70x+12y+4z=29x+5y+3z. So 41x+7y+z=0. Similarly, note that A=-C. That is, 6x+y-z=-35x-6y-2z. So 41x+7y+z=0. And B=-2A: 29x+5y+3z=-12x-2y+2z, or 41x+7y+z=0. So, because we've involved all three equations and discovered a common equation, there's dependence between the three equations and therefore no unique solution, but we can take one of the assumed variables and consider it as we would a constant. Let's call that constant c. The three equations can now be rewritten 6x+y=c-3, 29x+5y=6-3c, 35x+6y=3-2c. We only need two of these to find x and y. Take the first two and multiply the first by 5 and subtract the second: 30x+5y=5c-15 subtract 29x+5y=6-3c and we get x=8c-21. Substitute this in the first equation and we get 6(8c-21)+y=c-3, from which y=123-47c. We can check the answer by putting in c=0, for example, and checking out the three original equations after substituting for x and y.

### 3x+y=4 x+y=0

x + y =0 (solve this for y, you can also solve 3x + y=4 for y but in my opnion the first one is easier ) subtract x from both sides y = -x (substitute this one for y in the other equation) 3x - x = 4 (3x - 1x= 2x) 2x = 4 ( divide each side by 2) x = 2  ( this is one answer, plug this in to find the y value) 2 + y = 0 ( subtract 2 from each side) y = -2 x = 2   y = -2 is your answer check: 3(2) +(-2) = 4   6 + -2 = 4 this checks out -2 + 2 = 0  this also checks out the answer is correct.

### How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.

### find solution to the recurrence relation

Solving a Recurrence Relation Find the solution to the recurrence relation, a_n = 6a_(n-1) + 11a_(n-2) – 6a_(n-3), given a_0 = 20, a_1 = 5, a_2 = 15 Assuming a solution of the form a_n = c.r^n, then a_(n-1) = c.r^(n-1), a_(n-2) = c.r^(n-2), a_(n-3) = c.r^(n-3) Substituting for the above into the recurrence relation, c.r^n = 6c.r^(n-1) + 11c.r^(n-2) – 6c.r^(n-3) r^n = 6r^(n-1) + 11r^(n-2) – 6r^(n-3) 1 = 6r^(-1) + 11r^(-2) – 6r^(-3) r^3 = 6r^(2) + 11r^(1) – 6 r^3 – 6r^2 – 11r + 6 = 0 Solving the cubic equation above would give three root values for r, and a recurrence relation of the form a_n = c1.r1^n + c2.r2^n + c3.r3^n where the three constants, c1, c2, c3 are determined from the initial values for a_0, a_1 and a_2. The solutions for our cubic equation, r^3 – 6r^2 – 11r + 6 = 0, is r1 = -1.8256153, r2 = 0.4453157. r3 = 7.3802996 Our recurrence relation now becomes, a_n = c1*(-1.8256153)^n + c2*(0.4453157)^n + c3*(7.3802996)^n Using the initial values for a_0, a_1 and a_2, we get   a_0 = 20 = c1*(-1.8256153)^0 + c2*(0.4453157)^0 + c3*(7.3802996)^0 a_1 = 5 = c1*(-1.8256153)^1 + c2*(0.4453157)^1 + c3*(7.3802996)^1 a_2 = 15 = c1*(-1.8256153)^2 + c2*(0.4453157)^2 + c3*(7.3802996)^2   20 = c1 + c2 + c3  5 = c1*(-1.8256153 + c2*(0.4453157) + c3*(7.3802996) 15 = c1*(-1.8256153)^2 + c2*(0.4453157)^2 + c3*(7.3802996)^2   The last three simultaneous equations can now be solved for c1, c2 and c3. The values are: c1 = 1.990012, c2 = 17.9216148, c3 = 0.08837315 Our recurrence relation finally becomes, a_n = 1.990012*(-1.8256153)^n + 17.9216148*(0.4453157)^n + 0.08837315*(7.3802996)^n   Check Substituting for n = 0, 1, 2 in the final expression gives results for the initial values of a_0, a_1, a_2. n = 0:    19.99999995  (a_0 = 20) n = 1:    5.000000410   (a_1 = 5) n = 2:    15.00000017   (a_2 = 15)

### What will be their present agees?

Every problem has the answer built in. No-one is asking you to make up a process. Use the data given and simple equations, manipulate the equations, and the answer pops out. The sum of the ages of A and his father is 100. When A is as old as his father is now, he will be five times as old as his son B is now. B will be eight years older than A is now, when A is as old as his father is now. Start with the first statement. 1)   A + F = 100 The solution revolves around A being as old as his father is now, but x years from now 2)   F = A + x Substitute the value of F from equation 2 into equation 1 3)   A + (A + x) = 100 At that time in the future, A will be 5 times as old as his son is now 4)   A + x = 5B At that future date, the son will be 8 years older than A is right now 5)   B + x = A + 8 Solving that equation for B gives us the son's age right now 6)   B = A - x + 8 That takes care of the preliminaries. Substitute the value of B from equation 6 into equation 4 7) A + x = 5 (A - x + 8) Multiply through on the right side 8) A + x = 5A - 5x + 40 Add 5x to both sides, subtract A from both sides 9) 6x = 4A + 40 Divide both sides by 6 10) x = 2/3 A + 6 2/3 Substitute the value of x from equation 10 into equation 3 11) A + A + (2/3 A + 6 2/3) = 100 Combine the A terms, then subtract 6 2/3 from both sides 12) 2 2/3 A = 93 1/3 Multiply both sides by 3 13) 8A = 280 Divide both sides by 8 14) A = 35 We now know that A is 35 years old.   From equation 1, we find the age of A's father A + F = 100;    35 + F = 100;   F = 100 - 35;    F = 65 Putting that into equation 2, we find how far we are looking into the future F = A + x;    65 = 35 + x;    30 = x Substituting the appropriate values into equation 6, we find B's age B = A - x + 8;    B = 35 - 30 + 8;     B = 13   Check: 30 years in the future, B will be B + 30 = ??;    13 + 30 = 43 That should be 8 years older than A is now A + 8 = ??;      35 + 8 = 43 Done

student

### how to solve this system of equations using substitutioin: -4x+y=6, -5-y=21

- 4x + y = 6 - 5x - y = 21   To do this we need to rearrange one of the equations. It doesn't matter which one you choose to rearrange but I will choose the first.    - 4x + y = 6 Let's add '4x' to both sides of the equation so that we have an equation in the form y = ......   - 4x + 4x + y = 6 + 4x y = 6 + 4x   And so now we know what 'y' equals. We can substitute this into the second equation to help us find out what 'x' equals.        - 5x - y = 21 - 5x - (6 + 4x) = 21 - 5x - 6 - 4x = 21 - 9x - 6 = 21 - 9x - 6 + 6 = 21 + 6 - 9x = 27 x = 27/(-9) x = - 3   And so now we know what 'x' equals we can find out what 'y' equals by substituting x = - 3 into either of the original two equations.   I will use the equation,  - 4x + y = 6   , but you should also get the same answer if you choose the equation,  - 5x - y = 21  (a good way of checking if your answer is correct is to put the values for 'x' or 'y' into both equations and seeing if they both come up with the same answer )   So,  - 4x + y = 6 - 4*(- 3) + y = 6 12 + y = 6 12 - 12 + y = 6 - 12 y = - 6   This means that the solution to your set of equations is x = - 3 and y = - 6.

### Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell?

Shawn took in \$69.15 in two hours work on Saturday selling Belts for \$8.05 and earrings for \$4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ \$8.05 per belt. Let E be the number of earrings sold @ \$4.50 per pair. Total money made is P = \$69.15 Total money made is: B*\$8.05 + E*\$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear \$69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings

### find the general solution of the equation y"-y'-2y=x^2

find the general solution of the equation y"-y'-2y=x^2 Auxiliary eqn   (assuming a solution of the form y = A.e^(mx) ) m^2 – m – 2 = 0 (m – 2)(m + 1) = 0 m = 2, m = -1 1st solution: y1(x) = A.e^(2x) + B.e^(-x) Assume a solution of the form, y = Cx^2 + Dx + E Then Y’ = 2Cx + D Y’’ = 2C Substituting for these differentials into the original DE, y"-y'-2y=x^2, 2C – (2Cx + D) – 2(Cx^2 + Dx + E) = x^2 -2Cx^2 – 2(C + D)x + 2C - D – 2E = x^2 Comparing coefficients of the powers of x, -2C = 1 -2(C + D) = 0 2C – D - 2E = 0 Giving, C = - ½ , D = ½ , E = - ¾ Hence, y2(x) = - ½x^2 + ½x - ¾ The general solution is y = y1 + y2 Answer: y(x) = A.e^(2x) + B.e^(-x) - ½x^2 + ½x - ¾