HL Type 1 Maths Coursework

Maryam Allana

12 Brook

The aim of my report is to discover and examine the patterns found within the constants of the linear equations supplied. After acquiring the patterns I will solve the equations and graph the solutions to establish my analysis. Said analysis will further be reiterated through the creation of numerous similar systems, with certain patterns, which will aid in finding a conjecture. The hypothesis will be proven through the use of a common formula. (This outline will be used to solve both, Part A and B of the coursework)

Part A:

Equation 1: x+2y= 3

Equation 2: 2x-y=4

Equation 1 consists of three constants; 1, 2 and 3. These constants follow an arithmetic progression with the first term as well as the common difference both equaling to one. Another pattern present within Equation 1 is the linear formation. This can be seen as the equation is able to transformed into the formula ‘y = mx+c’ as it is able to form a straight line equation (shown below). Similar to Equation 1, Equation 2 also follows an arithmetic progression with constants of; 2, -1 and 4. It consists of a starting term of 2 and common difference of -3. As with Equation 1, Equation 2 is also linear forming the formula ‘y = mx+c’. When examining both Equation 1 and 2, an inverse pattern can be seen, where equation 1 is the inverse of equation 2 and vice versa. This can be proved by observing gradients of both the equations where equation 1 equals ‘y= -x/2 + 3/2’ and equation 2 equals ‘y= 2x+4’ (This is proven through technological means below)

x + 2y= 3- Equation 1

2x - y= 4- Equation 2

The equation must be solved simultaneously in order to acquire a solution. Therefore…

x + 2y= 3

4x - 2y= -8

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5x= -5/5 thus x= -1 and y= 2

Graph

The significance of the solution is the point of intersection with values of x= -1 and y=2 as seen with both the graphical and algebraic solutions. Furthermore both the lines intersect perpendicularly to each other (as seen above). Additionally the line of equation 1 is the inverse of the line of equation 2 and vice versa.

The creation of systems similar to that of equation one and two aid in discovering further patterns and establishing a conjecture (as seen below)

Equation 3: 5x +7y =9

Equation 4: 7x- 5y= -17

The constants of Equation 3 are 5, 7 and 9 which form an arithmetic sequence with a first term of 5 and a common difference of 2. The general formula for this particular sequence is un= 2n+3. Similarly the constants of Equation 4 are 7, -5 and -17 with a first term of 7 and a common difference of -12, which form an arithmetic sequence. The general formula for the sequence is un= -12n+19. In order to obtain the gradient, ‘y’ must be made the subject. In the case of the third equation y= (9-5x)/7 and in the case of the fourth equation y= (7x+17)/5. Therefore the gradient of the third equation is -5/7 and the gradient of the fourth equation is 7/5. When we multiply both the gradients we get -1, thus proving that both the lines are perpendicular to one another. Furthermore it proves that the fourth equation is the inverse of the third equation.

5x +7y =9 -Equation 3

7x- 5y= -17 -Equation 4

25x+ 35y =45

49x-35y= -119

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74x= -74 thus x= -1 and y= 2

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Equation 5: 7x+5y =3

Equation 6: 5x- 7y= -19

The constants of Equation 5 are 7, 5 and 3 which form an arithmetic sequence with a first term of 7 and a common difference of -2. The general formula for this particular sequence is un= -2n+9. Similarly the constants of Equation 6 are 5, -7 and -19 with a first term of 5 and a common difference of -12, which form an arithmetic sequence. The general formula for the sequence is un= -12n+17. In order to obtain the gradient, ‘y’ must be made the subject. In the case of the fifth equation y= (3-7x)/5 and in the case of...