AS-Level Maths: Core 1 for Edexcel C1.3 Algebra and functions 3 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 56 Boardworks Ltd 2005 Linear simultaneous equations Linear simultaneous equations Contents Simultaneous equations involving one linear and one quadratic equation Linear inequalities Quadratic inequalities Polynomials Examination-style questions 2 of 56

Boardworks Ltd 2005 Linear simultaneous equations An equation with two unknowns has an infinite number of solution pairs. For example: x+y=3 is true when x=1 and y=2 x = 4 and y=7 x = 6.4 and y = 3.4 We can represent the set of solutions graphically. The coordinates of every point on the line satisfy the equation.

3 of 56 and so on. y 3 0 x+y=3 3 x Boardworks Ltd 2005 Linear simultaneous equations Similarly, an infinite number of solution pairs exist for the equation yx=1 Again, we can represent the set of solutions graphically. There is one pair of values that satisfies both these equations simultaneously. This pair of values corresponds to

the point where the lines x + y = 3 and y x = 1 intersect: y yx=1 1 -1 0 x x+y=3 This is the point (1, 2). At this point x = 1 and y = 2. 4 of 56 Boardworks Ltd 2005 Linear simultaneous equations Two linear equations with two unknowns, such as x and y, can be solved simultaneously to give a single pair of solutions. When will a pair of linear simultaneous equations have no solutions? In the case where the lines corresponding to the equations are parallel, they will never intersect and so there are no solutions. Linear simultaneous equations can be solved algebraically using:

The elimination method, or The substitution method. The solution to the equations can be illustrated graphically by finding the points where the two lines representing the equations intersect. 5 of 56 Boardworks Ltd 2005 The elimination method If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example: Solve the simultaneous equations 3x + 7y = 22 and 3x + 4y = 10. Subtracting gives: 3x + 7y = 22 3x + 4y = 10 3y = 12 y=4 The terms in x have been eliminated. Substituting y = 4 into the first equation gives: 3x + 28 = 22

3x = 6 x = 2 6 of 56 Boardworks Ltd 2005 The elimination method We can check whether x = 2 and y = 4 solves both 3x + 7y = 22 3x + 4y = 10 by substituting them into the second equation. LHS = 3 2 + 4 4 = 6 + 16 = 10 = RHS So the solution is x = 2, y = 4. 7 of 56 Boardworks Ltd 2005 The elimination method Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example: 5x 2y = 31 4x + 3y = 11

Solve: 1 2 We need to have the same number in front of either the x or the y terms before adding or subtracting the equations. Call these equations 3 1 2 2 3 + 4 8 of 56 1 and 2 . 15x 6y = 93 + 8x + 6y = 22 23x 3

4 = 115 x=5 Boardworks Ltd 2005 The elimination method Substitute x = 5 in 1 : 5 5 2y = 31 25 2y = 31 2y = 6 x = 3 Check by substituting x = 5 and y = 3 into LHS = 4 5 + 3 3 2 : = 20 9 = 11 = RHS So the solution is x = 5, y = 3. 9 of 56

Boardworks Ltd 2005 The elimination method 10 of 56 Boardworks Ltd 2005 The substitution method Two simultaneous equations can also be solved by substituting one equation into the other. For example: Solve: y = 2x 3 2x + 3y = 23 Call these equations Substitute 1 into 2 1

and 2 1 2 . : 2x + 3(2x 3) = 23 2x + 6x 9 = 23 8x 9 = 23 8x = 32 x=4 11 of 56 Boardworks Ltd 2005 The substitution method Substituting x = 4 into 1 gives

y=243 y=5 Check by substituting x = 4 and y = 5 into 2 : LHS = 2 4 + 3 5 = 8 + 15 = 23 = RHS So the solution is x = 4, y = 5. 12 of 56 Boardworks Ltd 2005 The substitution method Solve: Call these equations 3x y = 9 8x + 5y = 1 1

and 2 1 2 . One of the equations needs to be arranged in the form x = or y = before it can be substituted into the other equation. Rearrange equation 1 : 3x y = 9 y = 9 3x y = 3x 9 13 of 56 Boardworks Ltd 2005 The substitution method

Now substitute y = 3x 9 into equation 2 : 8x + 5(3x 9) = 1 8x + 15x 45 = 1 23x 45 = 1 23x = 46 x=2 Substitute x = 2 into equation 1 to find the value of y: 6y=9 y = 3 y = 3 So the solution is x = 2, y = 3. 14 of 56 Boardworks Ltd 2005 One linear and one quadratic equation Contents Linear simultaneous equations Simultaneous equations involving one linear and

one quadratic equation Linear inequalities Quadratic inequalities Polynomials Examination-style questions 15 of 56 Boardworks Ltd 2005 One linear and one quadratic equation Suppose one of the equations in a pair of simultaneous equations is linear and the other is a quadratic of the form y = ax2 + bx + c. By considering the points where the graphs of the two equations might intersect we can see that there could be two, one or no pairs of solutions. 16 of 56 Boardworks Ltd 2005 One linear and one quadratic equation If the second equation contains terms in xy or y2 the shape

of the corresponding graph will not be a parabola but a circle, a hyperbola or an ellipse: A line and a circle A line and a hyperbola A line and an ellipse Again we can have two, one or no pairs of solutions. 17 of 56 Boardworks Ltd 2005 One linear and one quadratic equation When a pair of simultaneous equations contains one linear and one quadratic equation, we usually solve them by substitution. For example: Solve: Substituting equation 1 y = x2 + 1

y=x+3 1 into equation x2 + 1 = x + 3 2 2 gives Rearranging to give a quadratic equation of the form ax2 + bx + c = 0 gives x2 x 2 = 0 (x + 1)(x 2) = 0 x = 1 18 of 56 or x=2 Boardworks Ltd 2005 One linear and one quadratic equation We can substitute these values of x into one of the equations

y = x2 + 1 y=x+3 1 2 to find the corresponding values of y. It is easiest to substitute into equation When x = 1: y = 1 + 3 y=2 2 because it is linear. When x = 2: y=2+3 y=5 The solutions are x = 1, y = 2 and x = 2, y = 5. 19 of 56 Boardworks Ltd 2005 One linear and one quadratic equation

y We can demonstrate the solutions to y=x+3 y = x2 + 1 y=x+3 using a graph. y = x2 + 1 (2, 5) (1, 2) 0 x It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs, particularly when the solutions are not integers. 20 of 56 Boardworks Ltd 2005 One linear and one quadratic equation

Look at this pair of simultaneous equations: yx=1 x2 + y2 = 13 1 2 What shape is the graph given by x2 + y2 = 13? The graph of x2 + y2 = 13 is a circular graph with its centre at the origin and a radius of 13 . We can solve this pair of simultaneous equations algebraically using substitution. We can then sketch the graphs of the equations to demonstrate where they intersect. 21 of 56 Boardworks Ltd 2005 One linear and one quadratic equation yx=1 x2 + y2 = 13 Rearranging 1 gives

Substituting into 2 gives 1 2 y=x+1 x2 + (x + 1)2 = 13 x2 + x2 + 2x + 1 = 13 2x2 + 2x 12 = 0 x2 + x 6 = 0 (x + 3)(x 2) = 0 x = 3 22 of 56 or x=2 Boardworks Ltd 2005 One linear and one quadratic equation We can substitute these values of x into one of the equations y=x+1

x2 + y2 = 13 1 2 to find the corresponding values of y. It is easiest to substitute into equation When x = 3: 1 because it is linear. When x = 2: y = 3 + 1 y=2+1 y = 2 y=3 The solutions are x = 3, y = 2 and x = 2, y = 3. 23 of 56

Boardworks Ltd 2005 One linear and one quadratic equation Demonstrating these solutions graphically gives: x2 + y2 = 13 y y=x+1 (2, 3) 0 x (3, 2) The graphs intersect at the points (3, 2) and (2, 3). 24 of 56 Boardworks Ltd 2005 Using the discriminant In summary, to solve a pair of simultaneous equations where one equation is linear and the other is quadratic: Rearrange the linear equation so that one of the

variables is written in terms of the other. Substitute the linear equation into the quadratic equation to give a single equation of the form ax2 + bx + c = 0. We can find the determinant of this equation to find how many times the line and the curve will intersect. When b2 4ac > 0, there are two distinct points of intersection. b2 4ac = 0, there is one point of intersection (or two coincident points). The line is a tangent to the curve. b2 4ac < 0, there are no points of intersection. 25 of 56 Boardworks Ltd 2005 Using the discriminant Show that the line y 4x + 7 = 0 is a tangent to the curve y = x2 2x + 2. Call these equations and 2 . y 4x + 7 = 0 1 y = x2 2x + 2 2 y = 4x 7 Rearranging 1 gives 4x 7 = x2 2x + 2 Substituting into 2 gives x2 6x + 9 = 0 The discriminant = b2 4ac = (6)2 4(9)

= 36 36 =0 b2 4ac = 0 and so the line is a tangent to the curve. 26 of 56 1 Boardworks Ltd 2005 Linear inequalities Contents Linear simultaneous equations Simultaneous equations involving one linear and one quadratic equation Linear inequalities Quadratic inequalities Polynomials Examination-style questions 27 of 56 Boardworks Ltd 2005 Linear inequalities An inequality links two or more expressions with

the symbols: <, >, or . Inequalities are linear if the expressions they contain can be written in the form ax + b where a and b are constants. For example: 3x + 2 > 5 Solving this inequality involves finding the values of x that make the inequality true. In this example, the inequality is true when x > 1. The solution can be illustrated using a number line as follows: 3 28 of 56 2 1 0 1 2 3 4

5 Boardworks Ltd 2005 Solving linear inequalities Like an equation, we can solve an inequality by adding or subtracting the same value to both sides of the inequality sign. We can also multiply or divide both sides of the inequality by a positive value. For example: Solve 4x 7 > 11 2x. Add 7 to both sides: 4x > 18 2x Add 2x to both sides: 6x > 18 Divide both sides by 6: x>3 How could we check this solution? 29 of 56 Boardworks Ltd 2005 Checking solutions

To verify that is the solution to x>3 4x 7 > 11 2x substitute a value just above 3 into the inequality and then substitute a value just below 3. Substituting x = 4 into the inequality gives 4 4 7 > 11 2 4 16 7 > 11 8 9>3 This is true. Substituting x = 2 into the inequality gives 4 2 7 > 11 2 2 8 7 > 11 4 1>7 30 of 56 This is not true. Boardworks Ltd 2005 Multiplying or dividing by negatives Although most inequalities can be solved like equations we have to take great care when multiplying or dividing both sides

of an inequality by a negative value. The following simple inequality is true: 3 < 5 Look what happens if we multiply both sides by 1: 3 1 < 5 1 3 < 5 This is not true. To keep the inequality true we have to reverse the inequality sign: 3 > 5 31 of 56 Boardworks Ltd 2005 Multiplying or dividing by negatives Remember: when both sides of an inequality are multiplied or divided by a negative number the inequality is reversed. For example: 4 3x 10 3x 6 x 2 The inequality sign is reversed. We could also solve this type of inequality by collecting x terms on the right and reversing the inequality sign at the end. 4 3x 10

4 10 + 3x 6 3x 2 x x 2 32 of 56 Boardworks Ltd 2005 Solving combined linear inequalities The two inequalities 4x + 3 5 and 4x + 3 < 15 can be written as a single combined inequality: 5 4x + 3 < 15 We can solve this inequality as follows: Subtract 3 from each part: 2 4x < 12 Divide each part by 4: 0.5 x < 3 We can illustrate this solution on a number line as follows: 1 0.5 33 of 56

0 0.5 1 1.5 2 2.5 3 3.5 4 Boardworks Ltd 2005 Solving combined linear inequalities Some combined inequalities contain variables in more than one part. For example: x 2 3x + 2 2x + 7 Treat this as two separate inequalities: x 2 3x + 2

and 3x + 2 2x + 7 2 2x + 2 x+27 4 2x x5 2x We can write the complete solution as 2 x 5 and illustrate it on a number line as follows: 3 2 34 of 56 1 0 1 2 3 4 5

6 7 Boardworks Ltd 2005 Overlapping solutions Solve the following inequality and illustrate the solution on a number line: 2x 1 x + 2 < 7 Treating as two separate inequalities gives 2x 1 x + 2 and x+2<7 x12 x<5 x3 If x < 5 then it is also 3. The whole solution set is therefore given by x < 5. This can be seen on the number line: 3 2 35 of 56 1

0 1 2 3 4 5 6 7 Boardworks Ltd 2005 Solutions in two parts Solve the following inequality and illustrate the solution on a number line: 4x + 5 < 3x + 5 4x + 3 Treating as two separate inequalities gives 4x + 5 < 3x + 5 and 3x + 5 4x + 3

4x < 3x 5x+3 x<0 2x x2 We cannot write these solutions as a single combined inequality. The solution has two parts. 3 2 36 of 56 1 0 1 2 3 4 5 6

7 Boardworks Ltd 2005 Quadratic inequalities Contents Linear simultaneous equations Simultaneous equations involving one linear and one quadratic equation Linear inequalities Quadratic inequalities Polynomials Examination-style questions 37 of 56 Boardworks Ltd 2005 Quadratic inequalities Quadratic inequalities contain terms in both x2 and x. For example: x2 + x 6 0 (x + 3)(x 2) 0

Factorizing gives x2 + x 6 is equal to 0 when: x+3=0 x = 3 and x2=0 x=2 These values give the end points of the solution set: 5 4 38 of 56 3 2 1 0 1

2 3 4 5 Boardworks Ltd 2005 Quadratic inequalities To find the solution set we can substitute a value from each of the following three regions: region 1 5 4 region 2 3 2 1

0 region 3 1 2 3 4 5 into the original inequality x2 + x 6 0. When x = 4: 42 + 4 6 0 16 4 6 0 60 This is true and so values in region 1 satisfy the inequality. 39 of 56 Boardworks Ltd 2005 Quadratic inequalities

When x = 0 02 + 0 6 0 6 0 This is not true and so values in region 2 do not satisfy the inequality. 5 4 3 When x = 3 region 3 region 2 region 1 2 1 0 1 32 + 3 6 0 9+360 2

3 4 5 60 This is true and so values in region 3 satisfy the inequality. 40 of 56 Boardworks Ltd 2005 Quadratic inequalities We have shown that values in region 1 and region 3 satisfy the inequality x2 + x 6 0. region 1 5 4 region 2 3 2

1 0 region 3 1 2 3 4 5 4 5 We can show the complete solution set as follows: 5 4 3

2 1 0 1 2 3 So the solution to x2 + x 6 0 is: x 3 41 of 56 or x2 Boardworks Ltd 2005 Quadratic inequalities An alternative method for solving inequalities involves using graphs. For example: Solve x2 + x 3 > 4x + 1.

The first step is to rearrange the inequality so that all the terms are on one side and 0 is on the other. x2 3x 4 > 0 Sketching the graph of y = x2 3x 4 will help us to solve this inequality. The coefficient of x2 > 0 and so the graph will be -shaped. 42 of 56 Boardworks Ltd 2005 Quadratic inequalities Next, we find the roots by solving x2 3x 4 = 0. Factorizing gives (x + 1)(x 4) = 0 x = 1 or x=4 We can now sketch the graph. y

The inequality (4, 0) (1, 0) x2 3x 4 > 0 0 is true for the parts of the curve that lie above the x-axis. x So, the solution to x2 + x 3 > 4x + 1 is x < 1 43 of 56 or x>4 Boardworks Ltd 2005 Solving quadratic inequalities using graphs 44 of 56

Boardworks Ltd 2005 Polynomials Contents Linear simultaneous equations Simultaneous equations involving one linear and one quadratic equation Linear inequalities Quadratic inequalities Polynomials Examination-style questions 45 of 56 Boardworks Ltd 2005 Polynomials A polynomial in x is an expression of the form ax n + bx n 1 + cx n 2 +...+ px 2 + qx + r where a, b, c, are constant coefficients and n is a positive integer. The value of a is called the leading coefficient. Examples of polynomials include:

3x7 + 4x3 x + 8 x11 2x8 + 9x and 5 + 3x2 2x3. Polynomials are usually written in descending powers of x. They can also be written in ascending powers of x, especially when the leading coefficient is negative, as in the last example. 46 of 56 Boardworks Ltd 2005 Polynomials The degree, or order, of a polynomial is given by the highest power of the variable. A polynomial of degree 1 is called linear and has the general form ax + b. A polynomial of degree 2 is called quadratic and has the general form ax2 + bx + c. A polynomial of degree 3 is called cubic and has the general form ax3 + bx2 + cx + d. A polynomial of degree 4 is called quartic and has the general form ax4 + bx3 + cx2 + dx + e.

47 of 56 Boardworks Ltd 2005 Using function notation Polynomials are often expressed using function notation. For example, consider the polynomial function: f(x) = 2x2 7 We can use this notation to substitute given values of x. For example: Find f(x) when a) f(2) = 2(2)2 7 a) x = 2 b) x = t + 1 b) f(t + 1) = 2(t + 1)2 7 =87 = 2(t2 + 2t + 1) 7 =1 = 2t2 + 4t + 2 7 = 2t2 + 4t 5

48 of 56 Boardworks Ltd 2005 Adding and subtracting polynomials When two or more polynomials are added, subtracted or multiplied, the result is another polynomial. Polynomials are added and subtracted by collecting like terms. For example: Find f(x) = 2x3 5x + 4 a) f(x) + g(x) a) f(x) + g(x) and g(x) = 2x 4 b) f(x) g(x) b) f(x) g(x) = 2x3 5x + 4 + 2x 4

= 2x3 5x + 4 (2x 4) = 2x3 3x = 2x3 5x + 4 2x + 4 = 2x3 7x + 8 49 of 56 Boardworks Ltd 2005 Multiplying polynomials When two polynomials are multiplied together every term in the first polynomial must by multiplied by every term in the second polynomial. For example: f(x) = 3x3 2 and g(x) = x3 + 5x 1 f(x)g(x) = (3x3 2)(x3 + 5x 1) = 3x6 + 15x4 3x3 2x3 10x + 2 = 3x6 + 15x4 5x3 10x + 2 50 of 56

Boardworks Ltd 2005 Multiplying polynomials Sometimes we only need to find the coefficient of a single term. For example: Find the coefficient of x2 when x3 4x2 + 2x is multiplied by 2x3 + 5x2 x 6. We dont need to multiply this out in full. We only need to decide which terms will multiply together to give terms in x2. (x3 4x2 + 2x)(2x3 + 5x2 x 6) We have: 24x2 2x2 = 22x2 So, the coefficient of x2 is 22. 51 of 56 Boardworks Ltd 2005 Examination-style questions Contents Linear simultaneous equations Simultaneous equations involving one linear and one quadratic equation

Linear inequalities Quadratic inequalities Polynomials Examination-style questions 52 of 56 Boardworks Ltd 2005 Examination-style question 1 a) Solve the simultaneous equations x 2y = 2 x2 + 4y2 = 100 b) Interpret your solution geometrically. a) Label the equations Rearranging equation x 2y = 2 1 x2 + 4y2 = 100 2 1

x = 2 + 2y Substituting into equation 2 (2 + 2y)2 + 4y2 = 100 53 of 56 Boardworks Ltd 2005 Examination-style question 1 4 + 8y + 4y2 + 4y2 = 100 8y2 + 8y 96 = 0 y2 + y 12 = 0 (y + 4)(y 3) = 0 y = 4 Substituting into equation or y=3 1 When y = 4, x = 6 When y = 3, x = 8

b) The line x 2y = 2 crosses the curve x2 + 4y2 = 100 at the points (4, 6) and (3, 8). 54 of 56 Boardworks Ltd 2005 Examination-style question 2 a) Write an expression for the area A of the following rectangle: x2 x+2 b) If the area satisfies the inequality 5 < A < 12 find the range of possible values for a) x. A = (x + 2)(x 2) = x2 4 b) The range of possible values for x is given by 5 < x2 4 < 12 55 of 56 Boardworks Ltd 2005 Examination-style question 2

We have to solve 5 < x2 4 and then solve x2 4 < 12 5 < x2 4 x2 4 < 12 x2 9 > 0 x2 16 < 0 (x + 3)(x 3) > 0 (x + 4)(x 4) < 0 Sketching y = x2 16 Sketching y = x2 9 y y (3, 0) (3, 0) 0 (4, 0) (4, 0)

0 x x2 9 > 0 when x < 3 or x > 3 x x2 16 < 0 when 4 < x < 4 So the range of possible value for x is 3 < x < 4 (ignoring negative solutions) 56 of 56 Boardworks Ltd 2005