Solving Simultaneous Equations - Double Substitution Method

The following simultaneous equations may be quickly and easily solved by “double-substitution” method.

The conventional single substitution method is often times too lengthy and very prone to errors. Try your conventional method, before looking at  my "double substitution' method and you will know what I mean.

So, always have a quick and closer look at the simultaneous equations before you - if they are amenable to “double-substitution” method, then use it. (Now no longer in Form 2 or Form 3, so use a more "skillful method" where possible. hahaha  :))

1.        Solve the simultaneous equations

3x + y = 1

x² + 3xy + y² = 5 

                                                                                                        

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2.        Solve the simultaneous equations

              2x² + 9xy + 9y² = 0

              3x + 3y = 2

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The “Double-Substitution” Method:

Q 1)                             3x + y = 1                    …Eqn (1) 

x² + 3xy + y² = 5          …Eqn (2)

Fr. (1):                y = 1 – 3x                    …Eqn (3)

Fr. (2):                x² + y(3x + y) = 5         …Eqn (4)

Put (1) & (3) into (4):

                           x² + (1 – 3x)(1) = 5

                           x² + 1 – 3x = 5

                           x² – 3x – 4 = 0

                           (x + 1)(x – 4) = 0

                           x = -1                           or,        x = 4

Use (3):              y = 1 – 3(-1)                or,        y = 1 - 3(4)

                           y = 4                            or,        y = -11

Answer: x = -1, y = 4

               x = 4, y = -11

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Q2)                     2x² + 9xy + 9y² = 0      …Eqn (1)

3x + 3y = 2             …Eqn (2)

Fr. (2):                3y = 2 – 3x                  …Eqn (3)

Fr. (1):                2x² + 3y(3x + 3y) = 0 …Eqn (4)

Put (2) & (3) into (4):

                           2x² + (2 -3x)(2) = 0

                           2x² + 4 – 6x = 0

                           2(x² + 2 – 3x) = 0

                           2(x - 1)(x – 2) = 0

                           x = 1                            or,        x = 2

Use (3):              3y = 2 – 3(1)                or,        3y = 2 – 3(2)

                           y = -1/3                        or,        y = -4/3

Answer: x = 1, y = -1/3

               x = 2, y = -4/3

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You can also use 'double substitution' to solve part (c) of the question below easily. Try your conventional method first before you click here for my "double-substitution" method.

1)        Using cos(A + B) ≡ cos A cos B – sin A sin B,

(a)    show that

(i)             sin² Ө = (1/2)(1 – cos 2Ө),

(ii)           cos² Ө = (1/2)(cos 2Ө + 1)                                                            (4)

f(Ө) = 1 + 10 sin² Ө - 16 sin⁴ Ө .

(b)   Show that f(Ө) = 3 cos 2Ө – 2 cos 4Ө.                                               (4)

(c)    Solve the equation

1 + 10 sin² Ө^o - 16 sin⁴ Ө^o + 2 cos 4Ө^o = 0.25, for 0 ≤ Ө ≤ 180

             giving your solutions to 1 decimal place.                                               (4) 

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