Linear Equations in Two Variables

Linear Equations in Two Variables exercise

• Solve the linear equation.

  Hints

  The keyword each indicates multiplication.

  If you know the value of one of the variables, you can plug it into the equation and solve for the remaining variable.

  Solve equations using opposite operations. Just have a look at the example beside.

  Solution

  What do we know?

  Francis and Claire get $\$~5$ for each box of Snickerdoodles they sell and $\$~4$ for each box of Chocolate Party Rings they sell.

  • First, we assign $s$ to the unknown number of Snickerdoodle boxes sold. The price $\$~5$ each indicates multiplication: so $5s$ is the total amount of money made from selling Snickerdoodles.
  • In the same way, we assign $c$ to the unknown number of Chocolate Party Rings sold to get $4s$ as the total amount of money made from selling Chocolate Party Rings.
  • We add these amounts together, $5s+4c$, to get the total amount of money made from selling Snickerdoodles and Chocolate Party Rings.
  • Because Francis and Claire want to make $\$~500$ in total, we can establish the linear equation with two variables $5s+4c=500$.

  Now let's consider the two following situations:

  $\mathbf{c=0}$

  • Plugging $c=0$ into the equation gives us $5s=500$.
  • Dividing both sides by $5$ to isolate $s$ gives us $s=\frac{500}5=100$.

  So they have to sell $100$ packs of Snickerdoodles to reach their goal of $\$~500$.

  $\mathbf{s=0}$

  • Plugging $s=0$ into the equation gives us $4c=500$.
  • Dividing both sides by $4$ to isolate $c$ gives us $c=\frac{500}4=125$.

  So they have to sell $125$ packs of Chocolate Party Rings to reach their goal of $\$~500$.

• Complete the Table.

  Hints

  You have to solve the equation $5s+4c=600$ for each given value: Just look at the example beside for $s=100$.

  You see you first have to plug in the given value and than use opposite operations to isolate the unknown value.

  Solution

  Let's start with the solving of the equation for $c=0$:

  $\begin{array}{rcl} 5s+4(0) & = & ~600\\ 5s+0&=&~600\\ 5s&=&~600\\ \color{#669900}{\div5} & & \color{#669900}{\div5}\\ s & = & ~120 \end{array}$

  So, you have to proceed as follows:

  1. Plug the given value in the equation.
  2. Solve the equation using opposite operations as in the case of $c=0$ above.

  Let's try it one more time for $s=80$:

  $\begin{array}{rcl} 5(80)+4c & = & ~600\\ 400+4c&=&~600\\ \color{#669900}{-400} & &\color{#669900}{-400}\\ 4c & = & ~200\\ \color{#669900}{\div4} & & \color{#669900}{\div4}\\ c & = & ~50 \end{array}$

  Similarly you get

  • $s=60$ for given $c=75$
  • $c=100$ for given $s=40$
  • $s=20$ for given $c=125$

• Identify the solutions to the linear equations with two variables.

  Hints

  Plug in the given value into the equation and solve for the remaining variable using opposite operations.

  Check your solution by plugging in both the value you solved for as well as the given value into the equation.

  Solution

  We solve the unknown in each equation by plugging in the given value:

  $\mathbf{3s+2c=600}$ given that $s=100$:

  • $3(100)+2c=600$ or $300+2c=600$.
  • Subtracting $300$ from both sides leads to $2c=300$.
  • Lastly we divide both sides by $2$ to get $c=150$.

  $\mathbf{5s+3c=1200}$ given that $c=100$:

  • $5s+3(100)=1200$ or $5s+300=1200$.
  • Subtracting $300$ from both sides leads to $5s=900$.
  • We still have to divide both sides by $5$ to get $s=180$.

  $\mathbf{4s+6c=240}$ given that $s=15$:

  • $4(15)+6c=240$ or $60+6c=240$.
  • Subtracting $60$ from both sides leads to $6c=180$.
  • Almost done: divide both sides by $6$ to get $c=30$.

  $\mathbf{4s+10c=400}$ given that $c=20$:

  • $4s+10(20)=400$ or $4s+200=400$.
  • Subtracting $200$ from both sides leads to $4s=200$.
  • We still have to divide both sides by $4$ to get $s=50$.

• Determine the corresponding value which makes $3t+6p=750$ true.

  Hints

  For each $t$, there is precisely one $p$ and vice versa.

  You can plug a value into the equation from the top or bottom row of values and solve for the missing variable.

  Solving the equation above for $t$ leads to the linear equation $t=-2p+250$.

  Solution

  Now to solve the equation for each given value, we transform it once for isolating $t$ and another time isolating $p$.

  Let's start by solving for $t$:

  $\begin{array}{rcl} 3t+6p & = & ~750\\ \color{#669900}{-6p} & &\color{#669900}{-6p}\\ 3t & = & ~-6p+750\\ \color{#669900}{\div3} & & \color{#669900}{\div3}\\ t & = & ~-2p+250 \end{array}$

  Now we are in the position to plug in the given values for $p$:

  • $p=50$ leads to $t=-2(50)+250=-100+250=150$.
  • $p=75$ leads to $t=-2(75)+250=-150+250=100$.
  • $p=125$ leads to $t=-2(125)+250=-250+250=0$.

  Now let's take the original equation and solve for $p$:

  $\begin{array}{rcl} 3t+6p & = & ~750\\ \color{#669900}{-3t} & &\color{#669900}{-3t}\\ 6p & = & ~-3t+750\\ \color{#669900}{\div6} & & \color{#669900}{\div6}\\ p & = & ~-0.5t+125 \end{array}$

  We can now plug the given values for $t$ in:

  • $t=50$ leads to $-0.5(50)+125=-25+125=100$.
  • $t=150$ leads to $-0.5(150)+125=-75+125=50$.
  • $t=200$ leads to $-0.5(200)+125=-100+125=25$.

• Find $c$ when $s=60$ and when $s=80$.

  Hints

  The opposite operation of addition is subtraction:

  $\begin{array}{rcl} x+8 & = & ~30\\ \color{#669900}{-8} & &\color{#669900}{-8}\\ x & = & ~22 \end{array}$

  The opposite operation of multiplication is division:

  $\begin{array}{rcl} 8x& = & ~32\\ \color{#669900}{\div8} & &\color{#669900}{\div8}\\ x & = & ~4 \end{array}$

  You can check your solution by plugging the value you get for $c$ as well as the value you are give for $s$ into the equation, and see if you get $500$.

  For example, for $s=40$ and $c=75$:

  $5(40)+4(75)=200+300=500$ $~~~~~$✓

  Solution

  To solve the equation for $c$ with the given $s$ values, just plug each given value into the equation and isolate the variable $c$ using opposite operations:

  $\mathbf{s=60}$

  $\begin{array}{rcl} 5(60)+4c&=&~500\\ 300+4c&=&~500\\ \color{#669900}{-300} & &\color{#669900}{-300}\\ 4c & = & ~200\\ \color{#669900}{\div4} & & \color{#669900}{\div4}\\ c & = & ~50 \end{array}$

  $\mathbf{s=80}$

  $\begin{array}{rcl} 5(80)+4c&=&~500\\ 400+4c&=&~500\\ \color{#669900}{-400} & &\color{#669900}{-400}\\ 4c & = & ~100\\ \color{#669900}{\div4} & & \color{#669900}{\div4}\\ c & = & ~25 \end{array}$

• Sort the solutions to the linear equations with two variables.

  Hints

  Plug each given value into the corresponding equation and use opposite operations.

  The smallest solution is $40$ and the largest one is $300$.

  Solution

  Plug in the given value and use opposite operations to solve for the desired solution.

  1. $\mathbf{2u+v=240}$ with known $u=100$ leads to $200+v=240$. Subtracting $200$ from both sides results in $v=40$.
  2. Plug $v=100$ into the equation $\mathbf{4u+4v=600}$ to get $4u+400=600$. Next subtract $400$ from both sides to get $4u=200$ and finally divide both sides by $4$ to get $u=50$.
  3. For $\mathbf{4u+6v=1200}$, plug $u=180$ in. So we get $720+6v=1200$ and subtract $720$ from both sides to get $6v=480$. Lastly divide both sides by $6$ and we get that $v=80$.
  4. For $\mathbf{4v+5u=1200}$, we know that $u=120$. Plugging into this value leads to $4v+600=1200$. Next subtract $600$ from both sides to gain $4v=600$ and divide by $4$ to get $v=150$.
  5. The next equation is $\mathbf{3u+2v=700}$ with $u=100$. Plugging this value in gives us $300+2v=700$. Subtracting $300$ from both sides leads to $2v=400$ and dividing both sides by $2$ results in $v=200$.
  6. Just one more equation: $\mathbf{2u+3v=1200}$ with given $v=200$. Once again we plug the given value into $2u+600=1200$ and subtract $600$ from both sides to get $2u=600$. Finally, dividing both sides by $2$ leads to the solution $u=300$.