Amazing mathematics solving systems of equations substitution answer key

The solution to a system can usually be found by graphing, but graphing may not always be the most precise or the most efficient way to solve a system. 

Here is a system of equations:

\(\begin {cases} 3p + q = 71\\2p - q = 30 \end {cases}\)

The graphs of the equations show an intersection at approximately 20 for \(p\) and approximately 10 for \(q\).

Without technology, however, it is not easy to tell what the exact values are.

Instead of solving by graphing, we can solve the system algebraically. Here is one way. 

If we subtract \(3p\) from each side of the first equation, \(3p + q = 71\), we get an equivalent equation: \(q= 71 - 3p\). Rewriting the original equation this way allows us to isolate the variable \(q\). 

Because \(q\) is equal to \(71-3p\), we can substitute the expression \(71-3p\) in the place of \(q\) in the second equation. Doing this gives us an equation with only one variable, \(p\), and makes it possible to find \(p\).

\(\begin {align} 2p - q &= 30 &\quad& \text {original equation} \\ 2p - (71 - 3p) &=30 &\quad& \text {substitute }71-3p \text{ for }q\\ 2p - 71 + 3p &=30 &\quad& \text {apply distributive property}\\ 5p - 71 &= 30 &\quad& \text {combine like terms}\\ 5p &= 101 &\quad& \text {add 71 to both sides}\\ p &= \dfrac{101}{5} &\quad& \text {divide both sides by 5} \\ p&=20.2 \end {align}\)

Now that we know the value of \(p\), we can find the value of \(q\) by substituting 20.2 for \(p\) in either of the original equations and solving the equation.

\(\begin {align} 3(20.2) + q &=71\\60.6 + q &= 71\\ q &= 71 - 60.6\\ q &=10.4 \end{align}\)​​​​​​

\(\begin {align} 2(20.2) - q &= 30\\ 40.4 - q &=30\\ \text-q &= 30 - 40.4\\ \text-q &= \text-10.4 \\ q &= \dfrac {\text-10.4}{\text-1} \\ q &=10.4 \end {align}​​​​​​\)

The solution to the system is the pair \(p=20.2\) and \(q=10.4\), or the point \((20.2, 10.4)\) on the graph. 

This method of solving a system of equations is called solving by substitution, because we substituted an expression for \(q\) into the second equation.

The goal of this discussion is to look at one way to reason about the structure of a system of equations in order to determine the solution and then have students make their own reasoning about a different, but similar, system of equations.

Poll the class to see how many students agree with Tyler and how many students disagree with Tyler. If possible, invite students from each side to explain their reasoning. As students explain, it should come out that Tyler is correct and, if no student brings up the idea, make sure to point out that we can also visualize this by graphing the equations in the system and noting that the lines look parallel and will never cross.

In the previous activity, students noticed that if they knew what one variable was equal to, they could substitute that value or expression into another equation in the same problem. Point out that, in this problem, the expression \((x + y)\) is equal to 5 in the first equation. If, in the second equation, we replace \((x+y)\) with 5, the resulting equation is \(5 = 7\) which cannot be true regardless of the choice of \(x\) and \(y\).

Display the following system and ask students how many solutions they think it has and to give a signal when they think they know:

\(\displaystyle \begin{cases} 4x + 2y = 8\\2x + y = 5 \end{cases}\)

Once the majority of the class signals they have an answer, invite several students to explain their thinking. There are multiple ways students might use to reason about the number of solutions this system has. During the discussion, encourage students to use the terms coefficient and constant term in their reasoning. Introduce these terms if needed to help students recall their meanings. Bring up these possibilities if no students do so in their explanations:

• “Re-write the second equation to isolate the \(y\) variable and substitute the new expression into the first equation in order to find that the system of equations has no solutions.”
• “Notice that both equations are lines with the same slope but different \(y\)-intercepts, which means the system of equations has no solutions.”
• “Notice that \(4x+2y\) is double \(2x+y\), but 8 is not 5 doubled, so the system of equations must have no solution.”

Writing: MLR1 Stronger and Clearer Each Time. If time allows, use this routine to give students an opportunity to summarize the whole-class discussion. Display the prompt, “How can you use the structure of a system of equations to determine when there are no solutions?” Give students 3–4 minutes to write a response, then invite students to meet with 2–3 partners, to share and get feedback on their writing. Encourage listeners to ask their partner clarifying questions such as, “What do you notice about the coefficients and constant terms in the equations?”; “What do the equations tell us about the rates of change and initial values?; “Can you use the example of Tyler’s problem to explain that more?” Students can borrow ideas and language from each partner to strengthen their final product. This will help students solidify their understanding of how to use the structure of the system of equations to determine the number of solutions.
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