Summary Harder simultaneous equations : one linear and one quadratic worked examples

Harder Simultaneous equations

Example 1: Solve the following simultaneous equations.



x−5 y=2
2 2
x −xy−20 y =40


We will be given two equations, one linear (no powers) and one quadratic (with powers).

Step 1: Take the linear equation and make the letter x subject of the formula. We can make y
subject of the formula but we will end up having fractions, ideally we avoid this situation.

x=5 y +2


Step 2: Substitute x=5 y +2 in the quadratic equation, we expand brackets and we simplify and
solve accordingly.

(5 y +2)2−( 5 y+ 2 ) y −20 y 2=40

( 5 y +2 ) (5 y +2 ) − y ( 5 y +2 ) −20 y 2−40=0
2 2 2
25 y +10 y+10 y + 4−5 y −2 y−20 y −40=0
18 y−36=0
18 y=36
y=2


Step 3: Substitute y=2in the first (linear) equation to find the value of x

x=5 y +2
x=5 ( 2 ) +2
x=12


In this case we only ended up with one value of x and one value of y . Most of the time we will end
up with 2 values of x and 2 values of y as seen in the following examples. Make sure that you do the
subject of the formula and the substitutions correctly.




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