# In the present case by looking at the two equations

• No School
• AA 1
• 48

This preview shows page 29 - 32 out of 48 pages.

is to rearrange the equation that is easier to rearrange. In the present case by looking at the two equations it seems that equation (1) might be easier to rearrange. 3 1 6 2 6 2 3 1 2 4 6 4 6 2 (3) 2 2 4 2 2 4 4 x x x y y x y y x - + = = - = = - = - Now let us put (3) into (2): what we have to do here is to replace 3 1 2 2 y by x - : 3 1 9 3 3 9 4 3 3 4 3 3 4 3 4 3 2 2 2 2 2 2 2 ( 4 ) 3 2 3 9 8 3 6 9 2 2 2 2 5 15 15 5 15 3 (4) 2 2 5 x y x x x x x x y x x x x x x x - - = ⇒ - - - = ⇒ - - + = ⇒ - + = + × - + × + - + + = = - = ⇒ - = = = - - 14243 Now putting (4) in (3) we get: 3 1 3 3 6 ( 3) 3 2 2 2 2 2 y y y = - × - = + = = Therefore, the solution to our system of equations is 3 3 x and y = - = . Note : Now if you hate fractions you can write 3 9 1.5 4.5 2 2 x as x and as . So you will have: 7.5 4 1.5 3 4.5 2.5 7.5 3 2.5 x x x x - + = + ⇒ - = = = - - . According to (3), 3 1 1.5 0.5 2 2 y x y x = - = - . 3 we knowthat x = - 1.5 0.5( 3) 1.5 1.5 3 therefore y = - - = + = .

Subscribe to view the full document.

30 2.3.2 The elimination method In this method one has to subtract (or add) the equations from (or to) one another. Again let us try to solve our first example using the elimination method. 4 0 3 7 x y x y - = + = As before we have to label these equations: 4 0 (1) 3 7 (2) x y x y - = + = Now we need to match up the numbers in front of the ' ' x s or y s so that when we subtract (or add) one equation from (or to) the other, one of the variable disappear. Again we need to choose the easiest option. Looking at the system we can see that by adding (1) to (2) the ' y s disappear (or they are eliminated) 4 0 (1) 3 7 (2) 7 7 0 7 7 7 1 7 x y x y x x x - = + + = + = = = = We need now to replace x by its value in any of the equations (it does not matter which one; but always choose the easiest one to calculate). Looking at the system the easiest one to calculate is the first equation, (1). 4 0 we found that 1 4 1 0 4 0 4 x y and x y y y - = = × - = - = = As you can see we arrived to the same results as using the substitution method. We can also try our second example using the elimination approach. 2 4 6 4 3 3 x y x y + = - - = Again we have to label these equations: