Solving systems of equations with subscripts

Variables with subscripts are treated just like variables without them

In math and science you might meet a variable with a subscript. Don’t let them scare you, they’re just a way to keep track of variables that could be related to each other in some way.

What does a variable with a subscript look like?

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As an example, $t_1,\ t_2,\ t_3$ are all variables with subscripts. They could represent three different measurements of time for the same experiment. You read them as “time 1”, “time 2” and “time 3”, only it’s shorter to write them with the subscripts instead of writing them out.

Even though the variables can be related in some way, that doesn’t mean they have the same value. This means that if you’re solving systems of equations that have variables with subscripts, you’ll need to solve for each variable.

How to solve systems of equations with subscripted variables

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Finding the unique solution to the system of equations when the variables have subscripts

Example

Use any method to find the unique solution to the system of equations.

$R_1T_1=500$

$R_2=10R_1$

and

$R_2T_2=800$

$T_2=4-T_1$

Let’s come up with a plan. We know how to solve a pair of equations with two unknowns, so let’s see if we can rewrite $R_2T_2=800$ into an equation in terms of $R_1$ and $T_1$, so we can use it with the equation $R_1T_1=500$.

Since $T_2=4-T_1$ is already solved for $T_2$ and $R_2=10R_1$ is already solved for $R_2$, we can plug these two values into the third equation.

$R_2T_2=800$

$(10R_1)(4-T_1)=800$

Use the distributive property.

$10R_1(4)-10R_1(T_1)=800$

$40R_1-10R_1T_1=800$

We can divide everything by $10$ to make it a little bit easier.

$4R_1-R_1T_1=80$

When we put this together with the first equation $R_1T_1=500$, we can say

$4R_1-500=80$

$4R_1-500+500=80+500$

$4R_1=580$

$\frac{4R_1}{4}=\frac{580}{4}$

$R_1=145$

We know that $R_2=10R_1$, so we get

$R_2=10(145)$

$R_2=1,450$

We can also use $R_1$ to find $T_1$, with the equation $R_1T_1=500$.

$145(T_1)=500$

$\frac{145(T_1)}{145}=\frac{500}{145}$

$T_1=\frac{100}{29}$

We can use $R_2$ to find $T_2$ by using the equation $R_2T_2=800$.

$1,450(T_2)=800$

$\frac{1,450(T_2)}{1,450}=\frac{800}{1,450}$

$T_2=\frac{16}{29}$

Bringing all of our values together, we can say that

$(R_1, T_1)=\left(145, \frac{100}{29}\right)$

$(R_2, T_2)=\left(1,450, \frac{16}{29}\right)$

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Let’s look at a system of two equations with subscripts.

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Even though the variables can be related in some way, that doesn’t mean they have the same value.

Example

Solve the system of equations for $h_t$ and $x_t$.

$h_t=2x_t-4$

$h_t=\frac{1}{3}x_t+3$

Here both equations are equal to $h_t$, so we can set them equal to one another.

$2x_t-4=\frac{1}{3}x_t+3$

Let’s move the integers to the left.

$2x_t-4+4=\frac{1}{3}x_t+3+4$

$2x_t=\frac{1}{3}x_t+7$

Let’s move the $x_t$’s to the left.

$2x_t-\frac{1}{3}x_t=\frac{1}{3}x_t-\frac{1}{3}x_t+7$

$\frac{5}{3}x_t=7$

Multiply both sides by $3/5$.

$\frac{3}{5}\cdot\frac{5}{3}x_t=\frac{3}{5}\cdot7$

$x_t=\frac{21}{5}$

Now use the equation of your choice to solve for $h_t$. We’ll use $h_t=2x_t-4$.

$h_t=2x_t-4$

$h_t=2\left(\frac{21}{5}\right)-4$

$h_t=\frac{42}{5}-4$

$h_t=\frac{42}{5}-\frac{20}{5}$

$h_t=\frac{22}{5}$

So we can say

$(x_t, h_t)=\left(\frac{21}{5},\frac{22}{5}\right)$

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As you can see, if you have subscripts in a system of equations, simply use the approach you would normally use to solve it.

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