Week 17 in Math 10

What we learned in math 10 is how to solve systems of equations. It was pretty interesting to me that we can just get the value of the x and y using two equations. There are two ways to get the solution of a pair of equations, Substitution and Elimination.  Substitution is useful when there is a coefficient which value is 1. You can use substitution if you remember how to isolate something on an equation. We should choose the simpler equation (e.g x – 2y = 3) firstly and express one variable in terms of the other. After this step, we need to substitute the expression from the last step into the equation and solve the single variable equation. Lastly, substitute the solution into the equation in the first step to find the value of the other variable. We can solve every single of systems of equations with substitution, but it might be really complicated if there is not a simple equation, so we learned another way, Elimination. I prefer elimination to substitution because I think it is easy for all students in the class unless they do not know how to multiply. Using the method of elimination is very simple. If necessary, multiply each equation by a constant to obtain coefficients for x (or y) that are identical (except perhaps for the sign) and add or subtract the two equations to eliminate one of the variables. After this, solve the resulting equation to determine the value of one of the values. Lastly, substitute the solution the solution into either of the original equations to determine the value of the other variable. We have learned about those ways, However, sometimes, we cannot get a solution of the systems of equations. We must make sure it has zero solution if the slope is same on the equations, and it has infinite solutions if the equations are same.

For examples,

1. A rectangle is to be drawn with perimeter 64 cm. If the length is to be 14 cm more than the width, determine the area of the rectangle.

  • If the width is y and the length is x, the equations are 2x + 2y = 64 and x – y = 14, and the first equation is divided by 2, so the equations are x – y = 14 and x + y = 32 {= (2x + 2y = 64)/2} (it does not really matter to skip dividing steps)
  • Substitution 
  • I am going to isolate y on x + y = 32. x + y = 32 => y = – x + 32. We can just substitute the value of y on one of those equations. x – (– x + 32) = 14 => 2x – 32 = 14 => 2x = 46 => x = 23
  • Elimination 
  • I do not need to multiply any numbers on the equation and just add these two equations. (x + y = 32) + (x – y = 14) => 2x = 46 => x = 23.
  • We can just substitute the value of x on one of those equations. (23) + y = 32 => y = 32 – 23 => y = 9
  • So, the length is 23 cm, and the width is 9 cm.
  • The formula of the area of a rectangle is (length) × (width). (23) × (9) = 207
  • So, the area of the rectangle is 207 cm2

2.

  • The system of equations is 2x – 3y = – 5 and x + y = 5. I got the solutions in two different ways.
  • Substitution
  • I isolated x on the second equation, so the equation is x = – y + 5, and substitute the equation on x of the first equation. 2(– y + 5) – 3y = – 5. – 2y + 10 – 3y = – 5 => – 2y – 3y = – 5 – 10 => – 5y = – 15 => y = 3
  • Elimination 
  • The second equation is multiplied by – 2, – 2(x + y = 5) => – 2x – 2y = – 10, and substract by the first equation. (2x – 3y = – 5) + (– 2x – 2y = – 10) => – 5y = – 15 => y = 3
  • The last step is to substitute 3 on y. x + y = 5 => x + 3 = 5 => x = 5 – 3 => x = 2.
  • So, the solution is (2,3)

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