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

———————–
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: $\begin{array}{rcl} & 2x-y-3=0\text{ , }2{{x}^{2}}-10x+y+9=0\text{ (5 marks)} \\ & \\ \end{array}$ $\begin{array}{rcl} 2x-y-3&=&0\text{ } \\ y&=&2x-3\text{ }......\text{(1)} \\ 2{{x}^{2}}-10x+y+9&=&0\text{ }......(2) \\ \end{array}$

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

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

———————–
Substitute (1) into (2)
———————– $\begin{array}{rcl} p+p(3p-1)-2(3p-1)&=&0 \\ p+3{{p}^{2}}-p-6p+2&=&0 \\ 3{{p}^{2}}-6p+2&=&0 \end{array}$ $a=3\text{ , }b=-6\text{ , }c=2$ $\begin{array}{rcl} p&=&\frac{-(-6)\pm \sqrt{{{(-6)}^{2}}-4(3)(2)}}{2(3)} \\ &=&\frac{6\pm \sqrt{12}}{6} \\ &=&1.577\text{ or }0.423 \end{array}$

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From (1):
———————– $\begin{array}{rcl} k&=&3p-1 \\ k&=&3(1.577)-1\text{ or }k=3(0.423)-1 \\ k&=&3.731\text{ or }k=0.269 \\ \end{array}$ $\text{So p = 1}\text{.577, k = 3}\text{.731 and p = 0}\text{.423},\text{ k = 0}\text{.269}$

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