How do I Use Linear Equations to Solve Problems?
A linear equation, if graphed, will form a straight line. Linear equations do not use exponents, roots or division of one variable by another. So x = y is a linear equation, but x = y^2, x = √y and x/y = 2 are not linear equations. Linear equations are much easier to work with than nonlinear equations, and you can often solve simple problems in the real world using them.
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Describing Motion
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If something is moving at a constant speed, meaning it is not accelerating or decelerating, its position after any given length of time is described by a very simple linear equation, x = v * t, where "x" is your position, "v" is the speed and "t" equals time. Let's say you are moving at a steady 10 miles per hour and want to know how far you've gone after 3 hours. Just plug the values into the equation: x = 10 (3) = 30. After three hours you've covered 30 miles.
Describing Profit from Sales
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Suppose you are an entrepreneur who has just purchased 1,000 apples for $50. If you make 10 cents profit on each apple, you may want to find your total profit if you sell all the apples. Use the equation T = p * s - c, where T = total profit, p = profit on each sale, s = number of sales and c = your costs. Plugging the values into the equation, you have T= 0.1(1,000) - 50 = $50. If, on the other hand, you only sell 400 apples, the equation becomes T = 0.1(400) - 50 = - $10. You would be $10 in the hole.
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Manipulating Linear Equations
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Remember that an equation is an expression of equality, so you can perform any type of arithmetic operation on both sides of the equation. Suppose in the previous example you want to know how many apples you need to sell to break even. So now your unknown value is not T. You are setting that equal to zero to try to find "s," the total number of sales. Start with the same equation, T = p * s - c. Plug in the values that you know: 0 = 0.1(s) - 50. Now isolate the unknown variable. Start by adding 50 to both sides of the equation: 50 = 0.1(s). Now divide both sides by 0.1. You'll get 500 = s. You need to sell 500 apples to break even.
Simultaneous Linear Equations
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Now suppose that you are looking at a race between a tortoise and a hare. The hare moves at 10 miles per hour, while the tortoise only manages 2 miles per hour, but has a head start of 10 miles. The equation for the hare's position is x = 10t, while the equation for the tortoise is x = 2t + 10. The hare will catch up when both of them have the same position. That is to say, when 10t (the hare's position at any given time) = 2t + 10 (the tortoise's position). So 10t = 2t + 10. Subtract 2t from each side: 8t = 10. Divide both sides by 8: t = 10/8, which reduces to 1 ¼. The hare will catch the tortoise after 1 ¼ hours. Check your answer by plugging 1 ¼ back into either equation. You'll get the same result, 12.5 miles, for either the tortoise or hare. Graphically, this would be the point where the line describing the tortoise's position and the line describing the hare's position intersect.
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References
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