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Worked solution

Solve simultaneous equations without a graphing calculator

A graphing calculator is not to be used in answering this question.

Solve the simultaneous equations z+(i−1)w=0z+(i-1)w=0 and 3z+(1−i)w=13z+(1-i)w=1.

2008/MI/I/1

CJC · Example 8

Answer

w=1+i8w=\dfrac{1+i}{8}, z=14z=\dfrac14.

Compare your reasoning

Label the equations

z+(i−1)w=0(1),3z+(1−i)w=1(2).\begin{aligned}z+(i-1)w&=0 &&(1),\\3z+(1-i)w&=1 &&(2).\end{aligned}

From (1), z=−(i−1)w=(1−i)wz=-(i-1)w=(1-i)w (3). Substitute (3) into (2):

3(1−i)w+(1−i)w=1,4(1−i)w=1,w=14(1−i)⋅1+i1+i=18(1+i).\begin{aligned}3(1-i)w+(1-i)w&=1,\\4(1-i)w&=1,\\w&=\frac1{4(1-i)}\cdot\frac{1+i}{1+i}\\&=\frac18(1+i).\end{aligned}

This is (4). Substitute (4) into (3):

z=(1−i)w=(1−i)18(1+i)=18(1−i2)=14.\begin{aligned}z&=(1-i)w\\&=(1-i)\frac18(1+i)\\&=\frac18(1-i^2)\\&=\frac14.\end{aligned}

The substitution and conjugate multiplication give exact values without a graphing calculator.

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