How to solve systems of equations with subscripts

A subscript gives a variable a label. Once we know which label goes with which quantity, we can solve the system with the same substitution steps we already use.

Read the whole variable name

The variables $t_{1}$ , $t_{2}$ , and $t_{3}$ could represent three time measurements. They’re distinct variables, and the subscripts help us keep track of which measurement we mean. Every occurrence of the same name, such as $t_{1}$ , stands for the same unknown in the system.

Example: Substitute to reduce four unknowns to two

Let’s find $R_{1}$ , $T_{1}$ , $R_{2}$ , and $T_{2}$ in this system.

\begin{array}{l} R_{1} T_{1} = 500 \\ R_{2} = 10 R_{1} \\ R_{2} T_{2} = 800 \\ T_{2} = 4 - T_{1} \end{array}

Two equations already tell us what to put in for $R_{2}$ and $T_{2}$ . If we substitute those expressions into $R_{2} T_{2} = 800$ , the new equation uses only $R_{1}$ and $T_{1}$ .

\begin{array}{l} (10 R_{1}) (4 - T_{1}) = 800 \\ 40 R_{1} - 10 R_{1} T_{1} = 800 \\ 4 R_{1} - R_{1} T_{1} = 80 \end{array}

The first equation gives us the whole product $R_{1} T_{1} = 500$ . Plug in $500$ for that product, and we can solve for $R_{1}$ .

\begin{array}{l} 4 R_{1} - 500 = 80 \\ 4 R_{1} = 580 \\ R_{1} = 145 \end{array}

Use the result to find the remaining variables

Once we have $R_{1}$ , we can find $R_{2}$ from $R_{2} = 10 R_{1}$ and $T_{1}$ from $R_{1} T_{1} = 500$ . Then $T_{2} = 4 - T_{1}$ .

\begin{array}{l} R_{2} = 10 (145) = 1450 \\ T_{1} = \frac{500}{145} = \frac{100}{29} \\ T_{2} = 4 - \frac{100}{29} = \frac{16}{29} \end{array}

The last equation to check is $R_{2} T_{2} = 800$ : $1450 (\frac{16}{29}) = 800$ . The substitutions give one possible value for each variable, and those values satisfy all four equations.

Example: Equate two expressions for the same variable

Find $x_{t}$ and $h_{t}$ when $h_{t} = 2 x_{t} - 4$ and $h_{t} = \frac{1}{3} x_{t} + 3$ . The two right sides must be equal because they both equal $h_{t}$ .

\begin{array}{l} 2 x_{t} - 4 = \frac{1}{3} x_{t} + 3 \\ 2 x_{t} = \frac{1}{3} x_{t} + 7 \\ \frac{5}{3} x_{t} = 7 \\ x_{t} = \frac{21}{5} \end{array}

Plug that value into the first equation to find $h_{t}$ .

\begin{array}{l} h_{t} = 2 (\frac{21}{5}) - 4 \\ = \frac{42}{5} - \frac{20}{5} \\ = \frac{22}{5} \end{array}

The second equation gives $\frac{1}{3} (\frac{21}{5}) + 3 = \frac{22}{5}$ as well. So the solution is $(x_{t}, h_{t}) = (\frac{21}{5}, \frac{22}{5})$ .

Learn mathKrista KingJanuary 30, 2021math, learn online, online course, online math, algebra 2, algebra ii, algebra, systems of equations, subscripts, subscripted variables, variables with subscripts, solving systems, solving systems of equations, linear systems, solving linear systems

How to solve systems of equations with subscripts

Read the whole variable name

Example: Substitute to reduce four unknowns to two

Use the result to find the remaining variables

Example: Equate two expressions for the same variable

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