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Simultaneous equations Simultaneous equations are two equations with two unknowns. They are called simultaneous because they must both be solved at the same time. The first step is to try to eliminate one of the unknowns. Example Solve these simultaneous equations and find the values of x and y. Equation 1: 2x + y = 7 Equation 2: 3x - y = 8 Add the two equations to eliminate the ys: 2x + y = 7 3x - y = 8 ------------ 5x = 15 x = 3 Now you can put x = 3 in either of the equations. Substitute x = 3 into the equation 2x + y = 7: 6 + y = 7 y = 1 So the answers are x = 3 and y = 1 Solve the simultaneous equations: Equation 1: y - 2x = 1 Equation 2: 2y - 3x = 5 Rearranging Equation 1, we get y = 1 + 2x We can replace the 'y' in equation 2 by substituting it with 1 + 2x Equation 2 becomes: 2(1 + 2x) - 3x = 5 2 + 4x - 3x = 5 2 + x = 5 x = 3 Substituting x = 3 into Equation 1 gives us y - 6 = 1, so y = 7.

How to find y=mx+c and stuff like that 1. Find the equation of the line given 2 points EXAMPLE: Given points (-1,4) and (2,13), find the equation of the straight line that passes through them. solution: We know the equation will be of the form y=mx+c, so we just need to find the gradient, m and the y-intercept, c. These are simple enough exercises. Gradient, m = m = m = 9/3 m=3 so, we have y=3x+c now. To find c, we need to know x and y. Well, we do. We know 2 sets of x and y...the coordinates in the question, so just pick one pair and put them in the equation to find c...I will choose (-1,4): 4 = 3(-1) + c 4 = -3 + c --> c = 7 so the equation is y=3x + 7. Check now with your other pair of coordinates. Put the x-value in and see if the y-value you calculate matches that in the question: when x=2, y=3(2) + 7, so y=6+7 ie y=13, which is correct. So we KNOW the answer is correct. 2. Find the equation of the line given 1 point and the gradient EXAMPLE: Find the equation of the straight line that passes through point (3,4) and has a gradient of -5. solution: This is a lot simpler than the previous example, since we already know the gradient - no need to calculate it. Starting with y=mx+c, put in the information we know, namely, x,y and m... 4 = -5(3) + c 4 = -15 + c --> c = 19 so the equation is y=-5x+19 3. Find the gradient and y-intercept of a given straight line EXAMPLE: Find the gradient and y-intercept of the equation 3y + 12x - 7 = 0. solution: We need the equation in the y=mx+c form, so we need to re-arrange it. 3y + 12x - 7 = 0 --> add 7 to both sides 3y + 12x = 7 --> subtract 12x from both sides 3y = -12x + 7 --> now divide everything by 3 This is the same equation, but in the correct form, to tell us the gradient and y-intercept, ie gradient = -4 y-intercept = 7/3

How to do Pythagoras Pythagoras' Theorem relates the lengths of the sides in a right-angled triangle. A right-angled triangle has one angle of 90°. The side opposite the right angle is always the longest side, and is called the hypotenuse. For example you could have a right angle triangle with sidesof 3cm, 4cm and 5cm. The hypotenuse is the 5cm side because it is opposite the right-angle and it is the longest side. You draw squares on each of the sides of the triangle. We are going to examine the areas of each of the squares. Shorter sides The square on the 3cm side has an area of 3cm × 3cm = 9cm². The square on the 4cm side has an area of 4cm × 4cm = 16cm². Hypotenuse The square on the 5cm side has an area of 5cm × 5cm = 25cm². How they are related If you add together the areas of the squares on the two shorter sides, you get 25cm². This is the same as the area of the square on the hypotenuse.

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www.everythingmath.co.uk