Writing and Solving Linear Equations

Writing and Solving Linear Equations exercise

• Solve the linear equation using the sum of interior angles of a triangle theorem.

  Hints

  The interior angles of any triangle sum to $180^\circ$.

  Check your solution by adding all angles together; the sum must be $180^\circ$.

  • "Two times" indicates a factor of $2$.
  • "Three times" indicates a factor of $3$.

  Solution

  Each angle is unknown but they have a relationship to one another.

  The second angle is two times the first one and the third angle is three times the first one. Assigning the variable $x$ to the first angle, we can then express these relationships in the following way:

  • $m\angle 1=x$
  • The second angle is two times the first, or $m\angle 2=2x$
  • The third angle is three times the first, or $m\angle 3=3x$

  Using the fact that all three angles sum to $180^\circ$, we have the equation $x+2x+3x=180^\circ$.

  • Combining like terms gives us $6x=180^\circ$.
  • Lastly, we divide by $6$ to get $x=30^\circ$.

  We plug this value in for $x$ to get all angles:

  • $m\angle 1=30^\circ$
  • $m\angle 2=2\times 30^\circ=60^\circ$
  • $m\angle 3=3\times 30^\circ=90^\circ$

  The corresponding triangle has a right angle, the third one.

• Solve the equation using knowledge of linear equations and geometric properties.

  Hints

  The sum of all interior angles of any triangle is

  $m\angle 1 + m\angle 2 + m\angle 3=180^\circ$.

  The perimeter of a rectangle is given by $2l+2w$.

  Here you see the distributive property: you have to multiply the factor $a$ with each summand inside the parenthesis and add the resulting products.

  Solution

  We start with the equilateral triangle. All interior angles are equal $m\angle 1=m\angle 2=m\angle 3=x$ and thus $x+x+x=180^\circ$ or, combining like terms, $3x=180^\circ$. Last we divide by $3$ to get

  $m\angle 1=m\angle 2=m\angle 3=60^\circ$

  The angles of the next triangle have the following relationship:

  • $m\angle 1=x$
  • $m\angle 2=3x+3\frac13$
  • $m\angle 3=2\left(3x+3\frac13\right)$

  So we add all those terms: $x+3x+3\frac13+2\left(3x+3\frac13\right)=180^\circ$.

  1. We use the distributive property: $x+3x+3\frac13+6x+6\frac23=180^\circ$.
  2. Rearranging the terms leads to $x+3x+6x+3\frac13+6\frac23=180^\circ$.
  3. Now we combine the like terms to get $10x+10=180^\circ$.
  4. Next subtracting $10$ from both sides we get $10x=170^\circ$.
  5. Dividing by $10$ results in $x=17^\circ$.

  We still have to put $x=17^\circ$ in to get

  • $m\angle 1=17^\circ$
  • $m\angle 2=3\times 17^\circ+3\frac13^\circ=54\frac13^\circ$
  • $m\angle 3=2\left(54\frac13^\circ\right)=108\frac23^\circ$

  Finally, J.P. Irving needs our help for his rectangular fence with the length $l$ and the width $w$. What do we know? The length $l$ of the rectangle is $12~\text{cm}$ more than the width. Assigning $w=x$ we can conclude $l=x+12~\text{cm}$.

  The perimeter of a rectangle is given by $2l+2w$. Thus we can establish the following equation, which we solve using opposite operations:

  $\begin{array}{rclll} 2x+ 2(x+12) & = & ~44&|&\text{distributive property}\\ 2x+2x+24&=&~44&|&\text{combining like terms}\\ 4x+24&=&~44\\ \color{#669900}{-24} & & \color{#669900}{-24}\\ 4x & = & ~20\\ \color{#669900}{\div 4} & & \color{#669900}{\div 4}\\ x&=&~5 \end{array}$

  The desired width is $5~\text{cm}$ and the length is $l=5~\text{cm}+12~\text{cm}=17~\text{cm}$.

• Sort the steps needed to solve one variable linear equations in geometry.

  Hints

  Here you see an example for combining like terms:

  $3x+x+2x=6x$.

  All interior angles of a triangle sum to $180^\circ$.

  Solution

  Assigning $x$ to the first angle, we can represent each angle by

  • $m\angle 1=x=40^\circ$,
  • $m\angle 2=x+10^\circ$, and
  • $m\angle 3=2x+10^\circ$.

  Summing all the angles results in $180^\circ$. So we get the following equation:

  $x+x+10^\circ+2x+10^\circ=180^\circ$.

  • Combining like terms leads to $4x+20^\circ=180^\circ$.
  • Subtracting $20^\circ$ results in $4x=160^\circ$.
  • Dividing both sides by $4$ gives $x=40^\circ$.

  We plug this value in the other two angles to get

  • $m\angle 2=40^\circ+10^\circ=50^\circ$, and
  • $m\angle 3=2(40^\circ)+10^\circ=90^\circ$.

  The resulting triangle has a right angle.

• Solve the word problems using linear equations and geometric properties.

  Hints

  In a right triangle the (non right) remaining angles sum to $90^\circ$.

  Pay attention that Paul's fence only has three sides: a longer one and two shorter ones.

  The surface area of a cube with the side length $s$ is given by $6s^2$.

  In an isosceles triangle the base angles are equal.

  Solution

  Tyrese

  In a right triangle the remaining angles sum to $90^\circ$. This together with the fact that one angle is $30^\circ$ more than three times the other angle leads to $x+30^\circ+3x=90^\circ$.

  • Combining like terms ends in $4x+30^\circ=90^\circ$.
  • Subtracting $30^\circ$ from both sides gives $4x=60^\circ$.
  • Dividing both sides by $4$ gives $x=15^\circ$.

  So the one angle is $15^\circ$ and the other one $3\times 15^\circ+30^\circ=75^\circ$.

  $~$

  Paul

  Let $x$ be the shorter length. So $1.5x$ is the longer length. Now we add the lengths to get $2x+1.5x=140$.

  • First, we combine the like terms: $3.5x=140$.
  • Next, we divide both sides by $3.5$ to get $x=40$.

  Thus the shorter length is $40~ft$ and the longer one $1.5\times 40~ft=60~ft$.

  $~$

  Matt

  All sides of a cube have the same length. The surface area is $6s^2$.

  • The perimeter leads to $4s=8~in$ and dividing by $4$ results in $x=2~in$.
  • Putting this in the surface area formula finally gives us $6\times (2~in)^2=24~in^2$.

  $~$

  Carol

  Each isosceles triangle has two equal angles, the base angles which must be less than $90^\circ$.

  So we get $2x+90^\circ=180^\circ$.

  • We subtract $90^\circ$ to get $2x=90^\circ$.
  • Next, we divide both sides by $2$, which tells us that the angles are $45^\circ$ each.

• Find solutions to geometric problems.

  Hints

  The perimeter of a square is given by $4s$ and the area is given by $s^2$.

  The sum of all interior angles of any triangle is 180 degrees. So, if one angle is a right angle, then the sum of the two other angles is $90^\circ$.

  In a rectangle we can consider parallel lines of equal length.

  The area is given by $A=lw$.

  Solution

  Rectangle:

  • Three sides are given. The desired one has the same length as the corresponding parallel side. So it's $12~in$.
  • The area can be calculated by multiplying the side lengths: $A=(3~in)(12~in)=36~in^2$.

  $~$

  Square:

  • All sides have the same length, it's $s$.
  • The perimeter is given by the formula $24~in=4s$. Dividing by $4$ gives us the side length $=6~in$.
  • Squaring the side length leads to the area $A=(6~in)^2=36~in^2$.

  $~$

  Triangle:

  • The represented triangle has a right angle. Thus the sum of the other two angles is $90^\circ$.
  • This leads to the equation $x+x+20=90$.
  • Combining like terms and subtracting $20$ results in $2x=70$.
  • Last we divide by $2$ to get $x=35^\circ$.
  • The other angle is given by $x+20^\circ=55^\circ$.

• Decide which statements are true.

  Hints

  In an isosceles triangle we have two equal sides and two equal angles.

  At most one angle in a triangle is obtuse.

  For example, a $3:2:1$ ratio leads to $3x+2x+x=180^\circ$ and thus $x=30^\circ$.

  The three angles are

  • $m\angle 1=30^\circ$
  • $m\angle 2 = 60^\circ$
  • $m\angle 3 = 90^\circ$

  Solution

  An isosceles angle has two congruent angles. So either $m\angle 1=40^\circ=m\angle 2$ and thus $m\angle 3=100^\circ$, or $m\angle 1=40^\circ$ and $m\angle 20m\angle 3=70^\circ$. This means that the first statement is false, as there are two possibilities and not one given a right triangle with one angle of $40^\circ$.

  The area of a square is $s^2$. If we double $s$, we get $(2s)^2=4s^2$. This is four times the area of the starting square. And so the second statement is false.

  In a right triangle the remaining angles sum to $90^\circ$. If the smaller one (let it be $x$) is $18^\circ$ less than the larger one we get $x+x+18^\circ=90^\circ$. Subtracting $18^\circ$ and dividing by two after leads to $x=36^\circ$ and thus $54^\circ$ for the larger angle. The $5:3:2$ ratio leads to $5x+3x+2x=180^\circ$. Combining the like terms leads to $10x=180^\circ$. Next, we divide by $10$ to get $x=18^\circ$. So we get the following angles $2\times 18^\circ=36^\circ$, $3\times 54^\circ=36^\circ$ and $5\times 18^\circ=90^\circ$ So the third statement is true.

  An equilateral triangle has three equal angles, each of $60^\circ$. Cutting such a triangle in half would divide one of these three angles into two $30^\circ$ angles. So each half would have one angle of $60^\circ$ and one of $30^\circ$. So find the remaining angle of these two triangles, we subtract $30$ and $60$ from $180$: $180-30-60=90$. And so each triangles has angles of $30^\circ$, $60^\circ$, and $90^\circ$. So the last statement is true.