Simple equations with subscripts

Subscripts are those little numbers in the bottom right

Sometimes equations may have variables with subscripts (tiny numbers just after the variable). This is especially the case in science subjects such as chemistry or physics.

Why use subscripts instead of different variables? Well, the variables in science often represent something specific. For example, $P$ stands for pressure and in some equations you’ll have more than one pressure represented. In that case we use subscripts: $P_{1}$, $P_{2}$, $P_3$ and so on.

How do we handle subscripts in equations? Just like we do any other variable, but be careful to copy the subscripts down carefully as you work the equation and to not mistake the subscript for a number that has an operation such as multiplication attached to it.

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Solving equations with subscripted variables is just like solving any other equation

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Example problem for solving an equation that includes subscripts

Example

In chemistry we learn that the pressure and volume of a gas are related according to the equation $P_1 V_1-P_2 V_2=0$ where $P_1$ and $V_1$ are the original pressure and volume, and $P_2$ and $V_2$ are the new pressure and volume. If the original pressure is $1.4$ and the original volume is $210$, and if the new pressure is $28$, what is the new volume?

Start by plugging in what you do know (use parentheses when plugging numbers in). We know that $P_1=1.4$, $V_1=210$, and $P_2=28$, so we get

$P_1 V_1-P_2 V_2=0$

$(1.4)(210)-(28)V_2=0$

Simplify the left side using the order of operations.

$294-28V_2=0$

Solve by working backwards from the order of operations. $-28V_2$ is being subtracted, so add $28V_2$ to both sides.

$294-28V_2+28V_2=0+28V_2$

$294=28V_2$

Divide both sides by $28$ because division undoes multiplication.

$\frac{28V_2}{28}=\frac{294}{28}$

$V_2=10.5$

think of the subscript as part of the variable, and solve the equation like you would any other

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Let’s try another example of solving simple equations with subscripts.

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Example

Suppose a car travels at a speed of $50$ mph for $125$ miles, then speeds up and travels at a new constant speed for another $153$ miles. If the total time for the trip is $4.75$ hours, how fast does the car go during the second part of the trip?

Use the equation

$\frac{d_1}{v_1}+\frac{d_2}{v_2}=t$

where $d_1$ and $d_2$ are the first distance and the second distance, $v_1$ and $v_2$ are the first velocity (speed) and the second velocity, and $t$ is the time for the trip.

Start by plugging in what you do know, which is $d_1=125$, $v_1=50$, $d_2=153$, and $t=4.75$.

$\frac{d_1}{v_1}+\frac{d_2}{v_2}=t$

$\frac{125}{50}+\frac{153}{v_2}=4.75$

Simplify the left side using the order of operations.

$2.5+\frac{153}{v_2}=4.75$

Solve by working backwards from the order of operations. $153/v_2$ is being added by $2.5$ so subtract both sides by $2.5$.

$2.5-2.5+\frac{153}{v_2}=4.75-2.5$

$\frac{153}{v_2}=2.25$

Since the variable $v_2$ is in the denominator, multiply both sides by $v_2$ to bring it out of the denominator.

$\frac{153}{v_2}\cdot v_2=2.25\cdot v_2$

$153=2.25v_2$

Divide both sides by $2.25$.

$\frac{153}{2.25}=\frac{2.25v_2}{2.25}$

$68=v_2$

The car travels at $68$ mph for the second part of the trip.

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