Simple Equations

When solving a simple equation, it is helpful to think of the equation as a balance, with the equals sign $(=)$ being the fulcrum or center. Therefore, if an operation is performed on one side of the equation, the same must be done on the other side. Just as adding masses of $10\text{ kg}$ to both sides of a beam keeps it balanced, so does adding $10$ to both sides of an equation.

Solving an equation is the process of getting what we're looking for (or solving for) on one side of the $=$ and everything else on the other side. We're really sorting information.

If we're solving for $x$, we must get $x$ on one side by itself.

Note: A $\ _\square$, although not uniformly used, is preceded by a final solution to the equation.

Some equations involve only addition and/or subtraction.

If $x+8=12,$ what is $x?$

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To solve the equation $x + 8 = 12$, we must get $x$ by itself on one side (isolate $x$). Therefore, subtract $8$ from both sides to obtain

$\begin{aligned} x+8-8&=12-8\\ x+0&=4\\ x&=4. \ _\square \end{aligned}$

Solve for $y:$

$y-19=28.$

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To solve this equation, we must isolate $y$ on one side. To do this, simply move the $-19$ to the other side and change the sign to obtain

$\begin{aligned} y&=28+19\\ y&=47. \ _\square \end{aligned}$

For more on fractions and fraction arithmetic, see the article Fractions.

Some equations involve only multiplication and/or division. This is typically when the variable is already on one side of the equation, but there is either more than one of the variable, e.g. $2x$, or a fraction of the variable, e.g. $\frac{1}3x$ or $\frac{x}2$.

In the same manner that we respectively add or subtract a new number to or from both sides of an equation, we can multiply both sides of an equation by the same number, and the equation will be the same. This case is also true for dividing both sides by the same number as long as it is not zero.

Solve for $x:$

$3x=9.$

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To deal with the coefficient $3$ $($there is more than one of the variable $x,$ e.g. $3x),$ divide both sides of the equation by $3$ to obtain

$\begin{aligned} \frac{3x}3&=\frac93\\ 1\times{x}&=3\\ x&=3. \ _\square \end{aligned}$

If $\frac{y}7=28,$ what is $y?$

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To isolate $y$ on the left side, multiply both sides by $7$ to obtain

$\begin{aligned} 7\bigg(\frac{y}7\bigg)&=28\times7\\ \frac77\times y&=196\\ 1\times y&=196\\ y&=196. \ _\square \end{aligned}$

Solve for $x:$

$\frac34x=18.$

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To deal with the coefficient $\frac34$, multiply both sides by the reciprocal $\frac43$ to obtain

$\begin{aligned} \frac43\bigg(\frac34x\bigg)&=18\times\frac43\\ \frac{4\times3}{3\times4}x&=\frac{18\times4}3\\ \frac{12}{12}x&=\frac{72}3\\ 1\times x&=24\\ x&=24. \ _\square \end{aligned}$

to be accomplished

Main article: Multi-step equations

Sometimes equations require more than one step in order to solve for the desired value. It is helpful to think of the variable that we wish to isolate as trapped behind multiple chests, each with a different lock. Here, each lock represents an operation that can be unlocked with the appropriate operation $\big(+$ unlocks $-$, $-$ unlocks $+$, $\times$ unlocks $\div$, $\div$ unlocks $\times$, $^2$ unlocks $\small \sqrt{\text{ }}$, $\small \sqrt{\text{ }}$ unlocks $^2$, etc.$\big).$ With each apt operation applied to both sides of an equation, a lock is removed, thereby moving us closer to the desired variable.

What value of $x$ satisfies $\sqrt{3x+7}=4?$

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To solve for $x$,

$\begin{aligned} \sqrt{3x+7}&=4&\qquad(\text{write the equation})\\ 3x+7&=16&\qquad(\text{square both sides})\\ 3x&=9&\qquad(\text{subtract 7 from both sides})\\ x&=3. \ _\square&\qquad(\text{divide both sides by 3}) \end{aligned}$

Solve the following system of equations:

$\begin{aligned} 3x-2y&=0\\ 17x-7y&=13.\\ \end{aligned}$

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There are two ways to solve this problem.

Solution 1: Subsitution

We are given

$\begin{aligned} 3x-2y&=0&\quad(1)\\ 17x-7y&=13&\quad(2)\\ \end{aligned}$

and we need to find the values of $x$ and $y$. Deriving from equation $(1)$,

$3x-2y=0 \implies 3x=2y \implies \frac{3x}3=\frac{2y}3 \implies \frac33x=\frac{2y}3 \implies 1\times x=\frac{2y}3 \implies x=\frac{2y}3.$

We now have

$x=\frac{2y}3.\qquad(3)$

Substituting $(3)$ into $(2)$ gives us

$\begin{aligned} 17\bigg(\frac{2y}3\bigg)-7y&=13\\ \frac{17\times2y}3-7y&=13\\ \frac{34y}3-7y&=13.\\ \end{aligned}$

To subtract a whole number, e.g. $7y,$ from a fraction, e.g. $\frac{34y}3,$ we need to convert the whole number $7y$ to a similar fraction, a fraction with the same denominator as $\frac{34y}3$. To do this, initially ignore all variables, e.g. $y,$ and express $7$ as a quotient with divisor $3$. We get

$\begin{aligned} 7&=\frac{c}3\\ 7\times3&=\bigg(\frac{c}3\bigg)3\\ 21&=\frac33c\\ 21&=1\times c\\ 21&=c\\ c&=21.\\ \end{aligned}$

This means that $7=\frac{21}3$. Inserting $y$ back into the equation and expressing $7y$ as a similar fraction, we get

$\begin{aligned} \frac{34y}3-\bigg(\frac{21}3\bigg)y&=13\\ \frac{34y}3-\frac{21y}3&=13\\ \frac{34y-21y}3&=13\\ \frac{13y}3&=13\\ 3\bigg(\frac{13y}3\bigg)&=13\times3\\ \frac3313y&=39\\ 1\times13y&=39\\ 13y&=39\\ \frac{13y}{13}&=\frac{39}{13}\\ \frac{13}{13}y&=3\\ 1\times y&=3\\ y&=3. \ _\square \end{aligned}$

Substituting this into $(3)$ gives

$\begin{aligned} x&=\frac{2(3)}3\\ x&=\frac63\\ x&=2. \ _\square \end{aligned}$

Solution 2: Elimination

We are given

$\begin{aligned} 3x-2y&=0&\quad(1)\\ 17x-7y&=13, &\quad (2)\\ \end{aligned}$

and we need to find the values of $x$ and $y$. In eliminating a variable from the equation, we add the equation with a certain term of that variable to the equation with the inverse of the term, e.g. $(14x+2y=13)+(5x-2y=7)$ or $(2x-13z=9)+(6x+13z=31).$ Inverse means "equal opposite", so if the sign of a term is positive, the sign of its inverse is negative (indicated by inserting a $-$ beside the term). If the sign of a term is negative, the sign of its inverse is positive.

Let us choose to eliminate $y$. However, notice that $-2y$ and $-7y$ are not equal, but both are negative. Therefore, either equation $1$ or $2$ must be positive, and the other must remain negative. In order for the terms to have the same coefficient, we multiply each equation by the coefficient of the term in the other equation. That is,

$\begin{aligned} (1)&\times7\\ (2)&\times2\\\\ (3x-2y)7&=0\times7\\ (17x-7y)2&=13\times2\\\\ (3\times7)x-(2\times7)y&=0\times7\\ (17\times2)x-(7\times2)y&=13\times2\\\\ 21x-14y&=0\\ 34x-14y&=26. \end{aligned}$

Now, multiply only one of the two equations by $-1$ after multiplying each by said coefficients. That is,

$\begin{aligned} (21x-14y)(-1)&=0\times-1\\ 34x-14y&=26\\\\ \big(21\times(-1)\big)x-\big(14\times(-1)\big)y&=0\\ 34x-14y&=26\\\\ -21x-(-14y)&=0\\ 34x-14y&=26\\\\ -21x+14y&=0\\ 34x-14y&=26\\\\ -21x+14y&=0\\ +34x-14y&=26\\ ————&——\\ 13x+0y&=26\\\\ 13x+0&=26\\ 13x&=26\\ \frac{13x}{13}&=\frac{26}{13}\\ \bigg(\frac{13}{13}\bigg)x&=2\\ 1\times x&=2\\ x&=2. \end{aligned}$

Substituting this into $(1)$ gives us

$\begin{aligned} 3(2)-2y&=0\\ 6-2y&=0\\ 6&=2y\\ \frac62&=\frac{2y}2\\ 3&=\bigg(\frac22\bigg)y\\ 3&=1\times y\\ 3&=y\\ y&=3. \ _\square \end{aligned}$