GoFormative LaTeX Examples

If you want to work out the steps to solve an equation in \LaTeX see the example below.

Use the aligned environment to type a multi-line equation. You can add & to preserve the vertical alignment of the = by typing &= instead of just =. The end of each line in this environment should have a double slash \\. Note that the equation begins and ends with \begin{aligned} and \end{aligned} and do not use \\ for the last line. Use \boxed{} to draw a rectangle around text.

\begin{aligned} 3x + 2 &= x - 5 \\ 3x + 2 {\color{red} -x} &= x - 5 {\color{red}- x} &\text{ subtract $x$ from both sides} \\ 2x + 2 &= -5 \\ 2x + 2 {\color{red} - 2} &= -5 {\color{red} -2} &\text{Subtract $2$ from both sides} \\ 2x &= -7 \\ {\color{red}\dfrac{{\color{black}2x}}{2}} &= {\color{red}\dfrac{{\color{black}-7}}{2}} &\text{ Divide by 2 to isolate $x$}\\ x = -\dfrac{7}{2} \\[2pc] 3{\color{red}\left(\dfrac{-7}{2}\right)} + 2 &= {\color{red}\left(\dfrac{-7}{2}\right)} - 5 &\text{Substitute ${\color{red}\dfrac{-7}{2}}$ in for $x$ to check}\\ -\dfrac{21}{2} + \dfrac{4}{2} &= \dfrac{-7}{2} - \dfrac{10}{2} \\ -\dfrac{17}{2} &= -\dfrac{17}{2}\checkmark&\text{ So, $\boxed{x = -\dfrac{7}{2}}$ satistifies the original equation!} \end{aligned}

Show an alternate way to solve 3x + 2 = x - 5 use labels to show how your work connects to the process shown here.

For operations consider

2\cdot(3 \cdot 4)^{2 + 1} will produce output with \cdot for multiplication dot and a^{b} where b is the exponent.

\sqrt{2} and \sqrt[4]{128} produce radicals with and without indices.

For Geometry - you can type \overline{AB} \cong \overline{CD}. Or note that two triangles are congruent \triangle ABC \cong \triangle DEF. Angle comparisons can be written \angle A \cong \angle B or m\angle A = 30^\circ. For lines, rays and segments try

\overleftarrow{AB}, \overrightarrow{BC}, \overline{AB}, \overleftrightarrow{DE}. You can denote other shapes like a cirlce with \bigcirc A or \bigodot A and arcs with \overgroup{AB} or \widehat{AB}.

Explain why x = 4 is not a solution to the equation 3x + 1 = 14.

The quadratic formula states that for any quadratic equation of the form ax^{2} + bx + c = 0. The solutions will lie at x = \dfrac{-b \pm \sqrt{{\color{green}b^{2}-4ac}}}{2a}. Note that the parabola will have a line of symmetry at x = \dfrac{-b}{2a} and there are three cases to consider determined by the sign of the discriminant {\color{green}b^{2}-4ac}.

\begin{cases}b^2-4ac < 0 & \text{ 2 distinct imaginary roots} \\b^2 - 4ac = 0 &\text{ 1 double root} \\ b^2 -4ac > 0 &\text{ 2 distinct real roots}\end{cases}

Determine the type of roots for 2x^{2} - 5x + 3=-1.