GCSE: Further Simultaneous Equations Dr J Frost Last modified: 31 st August 2015.

Slides:

Advertisements

Similar presentations

Solve a System Algebraically

Advertisements

Revision - Simultaneous Equations II

GCSE: Curved Graphs Dr J Frost Last modified: 31 st December 2014 GCSE Revision Pack Reference: 94, 95, 96, 97, 98.

8.3 Solving Systems of Linear Equations by Elimination

Substitution Method September 9, 2014 Page in Notes.

Warm up 1. State the center and radius. 2. Write the following in standard form. Center (-9, 2) r = 8.

Drill Solve the linear system by substitution. 1.y = 6x – 11 -2x – 3y = x + y = 6 -5x – y = 21.

Algebra II w/ trig. Substitution Method: 1. Solve an equation for x or y 2. Substitute your result from step 1 into the other equation and solve for the.

7 = 7 SOLUTION EXAMPLE 1 Check the intersection point Use the graph to solve the system. Then check your solution algebraically. x + 2y = 7 Equation 1.

Thinking Mathematically Algebra: Graphs, Functions and Linear Systems 7.3 Systems of Linear Equations In Two Variables.

Lesson 9.7 Solve Systems with Quadratic Equations

Goal: Solve a system of linear equations in two variables by the linear combination method.

IGCSE Solving Equations Dr J Frost Last modified: 23 rd August 2015 Objectives: From the specification:

FP2 Chapter 4 – First Order Differential Equations Dr J Frost Last modified: 6 th August 2015.

GCSE: Algebra & Solving Linear Equations Dr J Frost Last modified: 24 th August 2013.

Structures 3 Sat, 27 November : :00 Solving simultaneous equations:  using algebra  using graphs.

Elimination Method: Solve the linear system. -8x + 3y=12 8x - 9y=12.

7.4. 5x + 2y = 16 5x + 2y = 16 3x – 4y = 20 3x – 4y = 20 In this linear system neither variable can be eliminated by adding the equations. In this linear.

Rewrite the numbers so they have the same bases i.e. 8 2 = (2 3 ) 2.

Solving Nonlinear Systems Section 3.5 beginning on page 132.

WARM UP GRAPHING LINES Write the equation in slope- intercept form and then Graph. (Lesson 4.7) 1.3x + y = 1 2.x + y = 0 3.y = -4 3.

Solve Systems of Equations Using Elimination Section 6.3.

Year 9: Simultaneous Equations

Warm-up. Systems of Equations: Substitution Solving by Substitution 1)Solve one of the equations for a variable. 2)Substitute the expression from step.

Algebra Review. Systems of Equations Review: Substitution Linear Combination 2 Methods to Solve:

5.1 Solving Systems of Linear Equations by Graphing

Objective: Use factoring to solve quadratic equations. Standard(s) being met: 2.8 Algebra and Functions.

SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE

Simultaneous Equations 1

6) x + 2y = 2 x – 4y = 14.

X.2 Solving Systems of Linear Equations by Substitution

Solving Quadratic Equations by the Complete the Square Method

GCSE: Quadratic Simultaneous Equations

GCSE: Quadratic Simultaneous Equations

Solve Systems of Equations by Graphing

GCSE: Non-Right Angled Triangles

3-2: Solving Systems of Equations using Substitution

3.2 Solve Linear Systems Algebraically

All pupils can solve inequalities algebraically and graphically

Solve a system of linear equation in two variables

3-2: Solving Systems of Equations using Substitution

Solving Systems of Equations using Substitution

Quadratic Simultaneous Equations

Lesson 7.1 How do you solve systems of linear equations by graphing?

Non-linear simultaneous equations

Non - Graphical Solution of Simultaneous Equations

3-2: Solving Systems of Equations using Substitution

Systems of Linear and Quadratic Equations

Solve Simultaneous Equations One Linear, one quadratic [Circle]

SIMULTANEOUS EQUATIONS 1

GCSE: Tangents To Circles

Unit 23 Algebraic Manipulation

Solving simultaneous linear and quadratic equations

Algebra 1 Mini Posters Systems of Linear and Quadratic Equations

Objectives Identify solutions of linear equations in two variables.

3-2: Solving Systems of Equations using Substitution

Solve the linear system.

Skills Check Graphing Circles & Writing Equations.

G Dear ©2009 – Not to be sold/Free to use

Chapter 8 Systems of Equations

Example 2B: Solving Linear Systems by Elimination

3-2: Solving Systems of Equations using Substitution

3-2: Solving Systems of Equations using Substitution

3-2: Solving Systems of Equations using Substitution

IGCSE Solving Equations

3-2: Solving Systems of Equations using Substitution

Solving Linear Systems by Graphing

Presentation transcript:

GCSE: Further Simultaneous Equations Dr J Frost Last modified: 31 st August 2015

Starter Solve the following simultaneous (linear) equations. 2x + 3y = 8 4x – y = -5 x = -0.5 y = 3 ?

RECAP :: Equation of a circle x y The equation of this circle is: x 2 + y 2 = 25  The equation of a circle with centre at the origin and radius r is: x 2 + y 2 = r 2 ?

Quickfire Circles 1 1 x 2 + y 2 = x 2 + y 2 = x 2 + y 2 = x 2 + y 2 = x 2 + y 2 = x 2 + y 2 = 36 ? ? ? ? ? ?

Motivation x y x – y = 2 x 2 + y 2 = 100 Given a circle and a line, we may wish to find the point(s) at which the circle and line intersect. How could we do this algebraically? STEP 1: Rearrange linear equation to make x or y the subject. x = y + 2 STEP 2: Substitute into quadratic and solve. (y + 2) 2 + y 2 = 100 y 2 + 4y y 2 = 100 2y 2 + 4y – 96 = 0 y 2 + 2y – 48 = 0 (y + 8)(y – 6) = 0 y = -8 or y = 6 STEP 3: Use either equation to find the values of the other variable. When y = -8, x = -6 When y = 6, x = 8 ? ? ?

Your Go y = x 2 – 3x + 4 y – x = 1 STEP 1: Rearrange linear equation to make x or y the subject. y = 1 + x STEP 2: Substitute into quadratic and solve for one variable. 1 + x = x 2 – 3x + 4 x 2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x = 1 or x = 3 STEP 3: Use either equation to find the values of the other variable. When x = 1, y = 2 When x = 3, y = 4 ? STEP 4 (OPTIONAL): Check that your pairs of values work. 2 = 1 2 – (3 x 1) + 4  4 = 3 2 – (3 x 3) + 4  ? ? ?

Exercises y = x 2 + 7x – 2 y = 2x – 8 x 2 + y 2 = 8 y = x + 4 y = x 2 y = x + 2 x 2 + y 2 = 5 x – 2y = 5 y = x 2 – x – 2 x + 2y = 11 y = x 2 – 2x – 2 x = 2y + 1 x = -3, y = -14 x = -2, y = -12 x = -2, y = +2 x = -3, y = -14 x = -2, y = -12 x = 1, y = -2 x + y = 1 x 2 + y 2 = 1 x 2 – y 2 = 15 2x + 3y = 5 x 2 – y 2 = 15 2x + 3y = 5 y = x 2 – 3x x = y – 9 x 2 – 4y + 7 = 0 y 2 – 6z + 14 = 0 z 2 – 2x – 7 = 0 [Source: BMO] x = 2 – √ 13, y = 11 – √ 13 X = 2 + √ 13, y = 11 + √ 13 x = 3, y = 4 x = -5/2, y = 27/4 x = 1, y = 0 x = 0, y = 1 x = -8, y = 7 x = 4, y = -1 x = 1, y = 2, z = 3 (Add equations, then complete the squares – you’ll end up with a sum of squares which must each be 0) x = 5, y = -3 x = 191/59, y = -255/59 ? ? ? ? ? ? ? ? ? ?  6 x = 3, y = 1 x = -1/2, y = -3/4 ?