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Linear Simultaneous Equations - Substitution Simplified Revision Notes

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2.3.2 Linear Simultaneous Equations - Substitution

The Substitution Method

This method involves rearranging one of the equations to make a single variable the subject, then substituting it into the other equation.

infoNote

Example: Solve the system

x2+y2=9(Equation A)x^2 + y^2 = 9 \quad \text{(Equation A)}

x=y5(Equation B)x = y - 5 \quad \text{(Equation B)}

  1. Rearrange Equation B to isolate xx:
  • Equation B: x=y5x = y - 5
  1. Substitute xx from Equation B into Equation A:
  • Substitute x=y5x = y - 5 into x2+y2=9x^2 + y^2 = 9:
  • (y5)2+y2=9(y - 5)^2 + y^2 = 9
  • y210y+25+y2=9y^2 - 10y + 25 + y^2 = 9
  • 2y210y+25=92y^2 - 10y + 25 = 9
  • 2y210y+16=02y^2 - 10y + 16 = 0
  • y25y+8=0y^2 - 5y + 8 = 0
  1. Solve the quadratic equation for yy:
  • Factorize the quadratic:
  • (y7)(y4)=0(y - 7)(y - 4) = 0
  • y=7ory=4y = 7 \quad \text{or} \quad y = 4
  1. Substitute back to find xx:
  • If y=7y = 7:
  • x=75=2x = 7 - 5 = 2
  • If y=4y = 4:
  • x=45=1x = 4 - 5 = -1
  1. Solutions:
  • (2,7)(2, 7)
  • (1,4)(-1, 4)

Calculator Method

  1. Use the quadratic equation solver to find the roots of the quadratic:
  • y=7y = 7
  • y=4y = 4 In a question that does not require full working, this is a valid method.
infoNote

Q3 (Jan 2013, Q4)

(i) Solve the simultaneous equations:

y=2x23x5y = 2x^2 - 3x - 5

10x+2y+11=010x + 2y + 11 = 0

  1. Substitute yy from the first equation into the second equation:
10x+2(2x23x5)+11=010x + 2(2x^2 - 3x - 5) + 11 = 0 10x+4x26x10+11=010x + 4x^2 - 6x - 10 + 11 = 0 4x2+4x+1=04x^2 + 4x + 1 = 0 (2x+1)2=0(2x + 1)^2 = 0 2x+1=02x + 1 = 0 x=12x = -\frac{1}{2}
  1. Substitute x=12x = -\frac{1}{2} back into the first equation to find yy:
y=2(12)23(12)5y = 2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) - 5 y=2(14)+325y = 2\left(\frac{1}{4}\right) + \frac{3}{2} - 5 y=12+325y = \frac{1}{2} + \frac{3}{2} - 5 y=25y = 2 - 5 y=3y = -3
  1. Solution:
(12,3)\left( -\frac{1}{2}, -3 \right)
infoNote

Q3 (Jan 2013, Q4)

(ii) What can you deduce from the answer to part (i) about the curve yy == 2x23x52x^2 - 3x - 5 and the line 10x+2y+11=010x + 2y + 11 = 0?

The line is a tangent to the curve at the point (12,3)\left( -\frac{1}{2}, -3 \right).

image

Note:

A tangent is a line that just touches the curve without cutting it.


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