GoFormative LaTeX Examples
star
star
star
star
star
Last updated almost 6 years ago
6 questions
Note from the author:
A set of demo items that you can use and adjust to incorporate LaTeX in formative items. Based on the KaTeX supported functions document.
1
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.
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.
1
Set-builder notation can be written \{x \mid x \in \mathbb{R}, -3 \leq x < 7\}. Note that other number sets can be written with \mathbb{letter}. So \mathbb{Q}, \mathbb{Z}, etc.Note that you cannot use these forms as correct answers...
Set-builder notation can be written \{x \mid x \in \mathbb{R}, -3 \leq x < 7\}. Note that other number sets can be written with \mathbb{letter}. So \mathbb{Q}, \mathbb{Z}, etc.
Note that you cannot use these forms as correct answers...
1
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 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.
1
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}.
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}.
1
Explain why x = 4 is not a solution to the equation 3x + 1 = 14.
Explain why x = 4 is not a solution to the equation 3x + 1 = 14.
1
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.
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.