# SPM Questions for Simultaneous Equations

At least one simultaneous equations question (5 marks) will be coming out in the SPM paper 2, section A.

1. Substitution (SPM 2005, SPM2007)
2. Using Formulae (SPM 2006, SPM 2009)

Substitution – SPM 2005 Paper 2 Question 1

Solve:
$x+\frac{1}{2}y=1\text{ and }{{y}^{2}}-10=2x$

Answer:

$\begin{array}{rcl} x+\frac{1}{2}y&=&1 \\ 2x+y&=&2 \\ 2x&=&2-y........(1) \\ {{y}^{2}}-10&=&2x ............(2) \end{array}$

———————–
Substitute (1) into (2)
———————–

$\begin{array}{rcl} & {{y}^{2}}-10=2-y \\ & {{y}^{2}}+y-12=0 \\ \end{array}$

$(y+4)(y-3)=0$
$y=-4\text{ or }y=\text{3}$

———————–
From (1):
———————–

$2x=2-y$
$2x=2-(-4)\text{ or }2x=2-(3)$
$2x=6\text{ or }2x=-1$
$x=3\text{ or }x=-\frac{1}{2}$

$\text{So x = 3, y = - 4 and x = -}\frac{1}{2},\text{ y = 3}$

SPM 2007 Paper 2 Question 1

Solve:

$\begin{array}{rcl} & 2x-y-3=0\text{ , }2{{x}^{2}}-10x+y+9=0\text{ (5 marks)} \\ & \\ \end{array}$

Answer:

$\begin{array}{rcl} 2x-y-3&=&0\text{ } \\ y&=&2x-3\text{ }......\text{(1)} \\ 2{{x}^{2}}-10x+y+9&=&0\text{ }......(2) \\ \end{array}$

———————–
Substitute (1) into (2)
———————–

$\begin{array}{rcl} & 2{{x}^{2}}-10x+2x-3+9=0 \\ & 2{{x}^{2}}-8x+6=0 \\ & {{x}^{2}}-4x+3=0 \\ & (x-3)(x-1)=0 \\ & x=3\text{ or }x=1 \\ \end{array}$

———————–
From (1):
———————–

$\begin{array}{rcl} y&=&2(3)-3\text{ or }y=2(1)-3 \\ y&=&3\text{ or }y=-1 \\ \end{array}$

$\text{So x = 3, y = 3 and x = }1,\text{ y = -1}$

Using Formulae – SPM 2006 Paper 2 Question 1

Solve:

$2x+y=1\text{ and }2{{x}^{2}}+{{y}^{2}}+xy=5$

Give your answers correct to three decimal places. (5 marks)

Answer:

$\begin{array}{rcl} 2x+y&=&1 \\ y&=&1-2x\text{ }......(1) \\ 2{{x}^{2}}+{{y}^{2}}+xy&=&5\text{ }......(2) \end{array}$

———————–
Substitute (1) into (2)
———————–

$\begin{array}{rcl} 2{{x}^{2}}+{{(1-2x)}^{2}}+x(1-2x)&=&5 \\ 2{{x}^{2}}+1-4x+4{{x}^{2}}+x-2{{x}^{2}}&=&5 \\ 4{{x}^{2}}-3x+1-5&=&0 \\ 4{{x}^{2}}-3x-4&=&0 \\ \end{array}$

$\text{a = 4, b = - 3, c = - 4}$

$\begin{array}{rcl} x&=&\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ &=&\frac{-(-3)\pm \sqrt{{{(-3)}^{2}}-4(4)(-4)}}{2(4)} \\ &=&\frac{3\pm \sqrt{73}}{8} \\ &=&1.443\text{ or -}0.693 \end{array}$

———————–
From (1):
———————–

$\begin{array}{rcl} y&=&1-2(1.443)\text{ or }y=1-2(-0.693) \\ y&=&-1.886\text{ or }y=2.386 \\ \end{array}$

$\text{So x = 1}\text{.443, y = -1}\text{.886 and x = -0}\text{.693},\text{ y = 2}\text{.386}$

SPM 2009 Paper 2 Question 1

Solve:

$k-3p=-1\text{ and }p+pk-2k=0$

Give your answers correct to three decimal places. (5 marks)

Answer:

$\begin{array}{rcl} k-3p&=&-1 \\ k&=&3p-1\text{ }......(1) \\ p+pk-2k&=&0\text{ }......(2) \end{array}$

———————–
Substitute (1) into (2)
———————–

$\begin{array}{rcl} p+p(3p-1)-2(3p-1)&=&0 \\ p+3{{p}^{2}}-p-6p+2&=&0 \\ 3{{p}^{2}}-6p+2&=&0 \end{array}$

$a=3\text{ , }b=-6\text{ , }c=2$

$\begin{array}{rcl} p&=&\frac{-(-6)\pm \sqrt{{{(-6)}^{2}}-4(3)(2)}}{2(3)} \\ &=&\frac{6\pm \sqrt{12}}{6} \\ &=&1.577\text{ or }0.423 \end{array}$

———————–
From (1):
———————–

$\begin{array}{rcl} k&=&3p-1 \\ k&=&3(1.577)-1\text{ or }k=3(0.423)-1 \\ k&=&3.731\text{ or }k=0.269 \\ \end{array}$

$\text{So p = 1}\text{.577, k = 3}\text{.731 and p = 0}\text{.423},\text{ k = 0}\text{.269}$

I hope this tutorial is useful to you. What do you think about this topic? Any comments and discussion are welcomed.