Transition Year Maths RULE


As a general rule a problem can be solved if there are as many, or more, equations as there are unknowns. If there are three unknowns, there must be at least three equations.


Example 11: Solve for x, y and z (equation 1)


x + 3y –2z


= 1


(equation 2) 4x – y – z = –1 (equation 3) 2x + 2y –3z = –3


Eliminate z from equation 1 and equation 2. x + 3y –2z = 1 (equation 1)


(Add)


–8x + 2y + 2z = 2 (multiply equation 2 by –2) –7x + 5y


= 3 (equation 4)


Eliminate z from equation 2 and equation 3. –12x + 3y + 3z = 3 (multiply equation 2 by –3) 2x + 2y –3z = –3 (equation 3) = 0 (equation 5)


(Add) –10x + 5y


Now use (equation 4) and (equation 5) to solve for x + y 7x –5y = –3 (multiply equation 4 by – 1) –10x + 5y = 0 (equation 5) –3x


(Add) =–3


⇒ x =1 ⇒ y = 2 (putting x = 1 back into equation 5) ⇒ z = 3 (putting x = 1 and y = 2 back into equation 1)


PROJECT 8.2 Solving for x, y, z and w.


Using a method similar to Example 11, solve for x, y, z, and w x + y + z + w =14


2x + 4y –3z + w = 9 3x –2y – z + 2w =6 –2x –3y + 2z + w =0


84


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