Solving a linear equation

Algebra, grade 7 Lesson 17 of 22Learning

Solving a linear equation

Solving an equation means finding all its solutions. Instead of guessing, we turn the equation, step by step, into simpler equations with the same solution, until we reach an equation like $\displaystyle {x}={5}$ – and the solution is written in it.

An equation as a balance scale

Whatever you do to one side, do to the other side too.

An equation is like a balance scale that is level: the two sides weigh the same. If we add the same number to both sides, or subtract the same number from both, they stay equal. So too if we multiply both sides by the same number, or divide both by the same number – as long as the number is not 0. The new equation has the same solution as the one before. Next to each step we write, after a slash, what we did in it.

Equations in one step

To get $\displaystyle x$ on its own, do the opposite operation: to undo adding, subtract; to undo multiplying, divide.

\begin{aligned}x+7&=12 &&/-7\\ x&=5\end{aligned}\qquad\qquad\begin{aligned}3x&=-12 &&/:3\\ x&=-4\end{aligned}

Equations in two steps

First get the term with

\displaystyle x

on its own, and only then

\displaystyle x

itself.

\begin{aligned}4x-5&=11 &&/+5\\ 4x&=16 &&/:4\\ x&=4\end{aligned}

Check: $\displaystyle {4\cdot4-5}={11}$ – true.

Exercise 1

Solve the equations.
Next to each step, write what you did in it.

Equations with brackets

When the equation has brackets, first simplify each side – with the distributive law and by collecting like terms – and only then solve:

\begin{aligned}3(x-2)+x&=10\\ 3x-6+x&=10\\ 4x-6&=10 &&/+6\\ 4x&=16 &&/:4\\ x&=4\end{aligned}

Exercise 2

Simplify the side with the brackets, then solve the equation.

The unknown on both sides

Bring all the terms with

\displaystyle x

to one side, and the numbers to the other.

When $\displaystyle x$ appears on both sides, subtract the same term from both so that $\displaystyle x$ is left on one side only:

\begin{aligned}5x+3&=2x+12 &&/-2x\\ 3x+3&=12 &&/-3\\ 3x&=9 &&/:3\\ x&=3\end{aligned}

A solution that is a fraction or a negative number

The solution can also be a negative number or a fraction:

\begin{aligned}2x+7&=1 &&/-7\\ 2x&=-6 &&/:2\\ x&=-3\end{aligned}\qquad\qquad\begin{aligned}3x-1&=4 &&/+1\\ 3x&=5 &&/:3\\ x&=\frac{5}{3}=1\frac{2}{3}\end{aligned}

Checking the solution

Always end with a check: substitute the solution into the original equation, into each side separately. Substituting $\displaystyle {x}={3}$ into the equation $\displaystyle {5x+3}={2x+12}$ , the first side comes out $\displaystyle {5\cdot3+3}={18}$ and the second $\displaystyle {2\cdot3+12}={18}$ – equal, so the solution is right. A check like this also catches a mistake in a sign.

Exercise 3

Solve, and check the solution by substitution.

Exercise 4

Dana solved the equation $\displaystyle {2(x-3)}={5x+6}$ and got $\displaystyle {x}={-4}$ .
Yoav solved the equation $\displaystyle {3x+2}={7}$ and got $\displaystyle {x}={3}$ .
Check by substitution: calculate both sides of Dana’s equation, and the first side of Yoav’s.
Which of them was right?
For the one who was wrong – solve the equation again.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: algebra, grade 7

Open as a page

Adding and subtracting signed numbers

The sum of two opposite numbers
$\displaystyle (-a)+a$ $\displaystyle =$ $\displaystyle 0$
Subtracting a number is adding its opposite
$\displaystyle a-b$ $\displaystyle =$ $\displaystyle a+(-b)$
Subtracting a negative number
$\displaystyle a-(-b)$ $\displaystyle =$ $\displaystyle a+b$

Multiplying and dividing signed numbers

Positive times negative – negative
$\displaystyle a\cdot(-b)$ $\displaystyle =$ $\displaystyle -(a\cdot b)$
Negative times negative – positive
$\displaystyle (-a)\cdot(-b)$ $\displaystyle =$ $\displaystyle a\cdot b$
Dividing: different signs – the quotient is negative
$\displaystyle (-a):b$ $\displaystyle =$ $\displaystyle -(a:b)$ $\displaystyle b\neq0$
Dividing: the same sign – the quotient is positive
$\displaystyle (-a):(-b)$ $\displaystyle =$ $\displaystyle a:b$ $\displaystyle b\neq0$

Powers and roots

A power – $\displaystyle a$ the base, $\displaystyle n$ the exponent: how many times $\displaystyle a$ is multiplied
$\displaystyle a^{n}$ $\displaystyle =$ $\displaystyle \underbrace{a\cdot a\cdot\ldots\cdot a}_{n}$
A number to the power 1 is the number itself
$\displaystyle a^{1}$ $\displaystyle =$ $\displaystyle a$
A negative base with an even exponent – the power is positive
$\displaystyle (-a)^{2}$ $\displaystyle =$ $\displaystyle a^{2}$
The square root – the positive number (or 0) whose square is the number under the root
$\displaystyle \sqrt{a^{2}}$ $\displaystyle =$ $\displaystyle a$ $\displaystyle a\ge0$

The laws of arithmetic and like terms

The commutative law of addition
$\displaystyle a+b$ $\displaystyle =$ $\displaystyle b+a$
The commutative law of multiplication
$\displaystyle a\cdot b$ $\displaystyle =$ $\displaystyle b\cdot a$
The associative law of addition
$\displaystyle (a+b)+c$ $\displaystyle =$ $\displaystyle a+(b+c)$
The associative law of multiplication
$\displaystyle (a\cdot b)\cdot c$ $\displaystyle =$ $\displaystyle a\cdot(b\cdot c)$
A number times itself – squared
$\displaystyle x\cdot x$ $\displaystyle =$ $\displaystyle x^{2}$
Collecting like terms – add the coefficients; the letters stay
$\displaystyle ax+bx$ $\displaystyle =$ $\displaystyle (a+b)x$

The distributive law

The distributive law
$\displaystyle a(b+c)$ $\displaystyle =$ $\displaystyle ab+ac$
The distributive law over a difference
$\displaystyle a(b-c)$ $\displaystyle =$ $\displaystyle ab-ac$
A minus before brackets – every term inside changes its sign
$\displaystyle -(a-b)$ $\displaystyle =$ $\displaystyle -a+b$

Distance, speed and time

$\displaystyle s$ the distance, $\displaystyle v$ the speed, $\displaystyle t$ the time
$\displaystyle s$ $\displaystyle =$ $\displaystyle v\cdot t$