•  A linear equation is an equation which has a power of \(1\) for each variable.

•  A non-linear equation is an equation which has at least one variable whose power is not \(1\).

• Solving simultaneous equations means finding the values of variables that satisfy those equations.

• \(3x+7y=81\)

• \(4x^2+5y^2=90\)
• \(\dfrac{1}{x}+\dfrac{2}{y}=15\)

Solve the following simultaneous equations using the substitution method.

\(\begin{aligned} 2x+y&=4\\ y^2+5&=4x \end{aligned}\)

Label all the equations.

\(\begin{aligned} 2x+y&=4 \quad\,\,\, \cdots \boxed{1} \\ y^2+5&=4x \quad \cdots \boxed{2} \end{aligned}\)

From \(\boxed{1}\), 

\(\begin{aligned} 2x&=4-y \\ x&=\dfrac{4-y}{2} \quad \cdots \boxed{3}. \end{aligned}\)

Substitute \(\boxed{3}\) into \(\boxed{2}\).

\(\begin{aligned} y^2+5&=4\left(\dfrac{4-y}{2}\right) \\ y^2+5&=8-2y \\ y^2+2y-3&=0 \\ (y+3)(y-1)&=0 \end{aligned}\)

\(y=-3\) or \(y=1\).

Substitute \(y=-3\) and \(y=1\) into \(\boxed{3}\).

\(\begin{aligned} x&=\dfrac{4-(-3)}{2} \\ &=\dfrac{7}{2} \end{aligned}\) or \(\begin{aligned} x&=\dfrac{4-1}{2} \\ &=\dfrac{3}{2}. \end{aligned}\)

Thus, \(x=\dfrac{7}{2}\), \(y=-3\) and \(x=\dfrac{3}{2}\), \(y=1\) are the solutions to these simultaneous equations.

Solve the following simultaneous equations using the elimination method.

\(\begin{aligned} 2x+y&=4 \\ x^2+2xy&=3 \end{aligned}\)

\(\begin{aligned} 2x+y&=4 \quad\,\,\ \cdots \boxed{1} \\ x^2-2xy&=3 \quad\,\,\ \cdots \boxed{2} \\ \boxed{1} \times 2x: 4x^2+2xy&=8x \quad \cdots \boxed{3} \\ \boxed{2}+\boxed{3}: \quad\quad\,\,\,\ 5x^2&=3+8x \\ 5x^2-8x-3&=0 \\ x&=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &=\dfrac{8 \pm \sqrt{(-8)^2-4(5)(-3)}}{2(5)} \end{aligned}\)

\(x=1.9136\) or \(x=-0.3136\).

Substitute \(x=1.9136\) into \(\boxed{1}\).

\(\begin{aligned} 2(1.9136)+y&=4 \\ 3.8272+y&=4 \\ y&=0.1728. \end{aligned}\)

Substitute \(x=-0.3136\) into \(\boxed{1}\).

\(\begin{aligned} 2(-0.3136)+y&=4 \\ -0.6272+y&=4 \\ y&=4.6272. \end{aligned}\)

Thus, \(x=1.9136\), \(y=0.1728\) and \(x=-0.3136\), \(y=4.6272\) are the solutions to these simultaneous equations.

•  A linear equation is an equation which has a power of \(1\) for each variable.

•  A non-linear equation is an equation which has at least one variable whose power is not \(1\).

• Solving simultaneous equations means finding the values of variables that satisfy those equations.

• \(3x+7y=81\)

• \(4x^2+5y^2=90\)
• \(\dfrac{1}{x}+\dfrac{2}{y}=15\)

Solve the following simultaneous equations using the substitution method.

\(\begin{aligned} 2x+y&=4\\ y^2+5&=4x \end{aligned}\)

Label all the equations.

\(\begin{aligned} 2x+y&=4 \quad\,\,\, \cdots \boxed{1} \\ y^2+5&=4x \quad \cdots \boxed{2} \end{aligned}\)

From \(\boxed{1}\), 

\(\begin{aligned} 2x&=4-y \\ x&=\dfrac{4-y}{2} \quad \cdots \boxed{3}. \end{aligned}\)

Substitute \(\boxed{3}\) into \(\boxed{2}\).

\(\begin{aligned} y^2+5&=4\left(\dfrac{4-y}{2}\right) \\ y^2+5&=8-2y \\ y^2+2y-3&=0 \\ (y+3)(y-1)&=0 \end{aligned}\)

\(y=-3\) or \(y=1\).

Substitute \(y=-3\) and \(y=1\) into \(\boxed{3}\).

\(\begin{aligned} x&=\dfrac{4-(-3)}{2} \\ &=\dfrac{7}{2} \end{aligned}\) or \(\begin{aligned} x&=\dfrac{4-1}{2} \\ &=\dfrac{3}{2}. \end{aligned}\)

Thus, \(x=\dfrac{7}{2}\), \(y=-3\) and \(x=\dfrac{3}{2}\), \(y=1\) are the solutions to these simultaneous equations.

Solve the following simultaneous equations using the elimination method.

\(\begin{aligned} 2x+y&=4 \\ x^2+2xy&=3 \end{aligned}\)

\(\begin{aligned} 2x+y&=4 \quad\,\,\ \cdots \boxed{1} \\ x^2-2xy&=3 \quad\,\,\ \cdots \boxed{2} \\ \boxed{1} \times 2x: 4x^2+2xy&=8x \quad \cdots \boxed{3} \\ \boxed{2}+\boxed{3}: \quad\quad\,\,\,\ 5x^2&=3+8x \\ 5x^2-8x-3&=0 \\ x&=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &=\dfrac{8 \pm \sqrt{(-8)^2-4(5)(-3)}}{2(5)} \end{aligned}\)

\(x=1.9136\) or \(x=-0.3136\).

Substitute \(x=1.9136\) into \(\boxed{1}\).

\(\begin{aligned} 2(1.9136)+y&=4 \\ 3.8272+y&=4 \\ y&=0.1728. \end{aligned}\)

Substitute \(x=-0.3136\) into \(\boxed{1}\).

\(\begin{aligned} 2(-0.3136)+y&=4 \\ -0.6272+y&=4 \\ y&=4.6272. \end{aligned}\)

Thus, \(x=1.9136\), \(y=0.1728\) and \(x=-0.3136\), \(y=4.6272\) are the solutions to these simultaneous equations.