First-degree equations
On one pan of a balance there are 3 mystery boxes and 2 weights of 1 kg; on the other, 11 weights of 1 kg. The balance is level. How much does each box weigh? Without realising it, you have just set up an equation: $3x + 2 = 11$. Solving it means finding the weight of the box without unbalancing the scales.
What an equation is
Don't confuse an equation with an identity: $2(x + 1) = 2x + 2$ is true for any $x$, so it is an identity, not an equation. And $3x + 2$ (with no $=$) is just an expression.
Checking a solution is easy: you substitute and see whether both sides give the same result. For $x = 3$: $3 \cdot 3 + 2 = 11$ ✓. For $x = 4$: $3 \cdot 4 + 2 = 14 \neq 11$ ✗.
In mathematics, it is essential to distinguish between an algebraic expression and an equation. Look at the following list and identify which ones are linear equations.…
In mathematics, it is essential to distinguish between an algebraic expression and an equation. Look at the following list and identify which ones are linear equations. For those that are, state their left-hand side and their unknown:
- a) $3x + 5$
- b) $2x - 8 = 10$
- c) $x^2 - 4 = 0$
- d) $5 + 3 = 8$
- e) $\frac{y}{2} + 1 = 7$
Checking whether a number is a solution to an equation involves substituting the unknown and seeing if the equality holds. Given the following set of values:…
Checking whether a number is a solution to an equation involves substituting the unknown and seeing if the equality holds. Given the following set of values: $\{-2, 0, 3\}$, state with reasoning which one of them is the solution to the equation:
$5x - (x + 4) = 8$
Solving: the balance
The golden rule: whatever you do to one side, do it to the other side too. That way the balance stays level and the equality stays true. Solve the balance:
That idea is what in class is called transposing terms:
Solve the following basic equations using transposition of terms (moving terms to the opposite side by changing their operation): a) $x + 15 = 42$ b) $x - 7 = -12$ c)…
Solve the following basic equations using transposition of terms (moving terms to the opposite side by changing their operation):
- a) $x + 15 = 42$
- b) $x - 7 = -12$
- c) $10 - x = 4$
- d) $25 = x + 18$
A pair of scales is balanced. On the left pan there are $3$ identical boxes (of weight $x$) and a weight of $2$ kg. On the right pan there is a weight of $14$ kg. Set up…
A pair of scales is balanced. On the left pan there are $3$ identical boxes (of weight $x$) and a weight of $2$ kg. On the right pan there is a weight of $14$ kg.
Set up the equation that represents this balance and calculate how much each box weighs.
Brackets and denominators
Real equations usually come "tangled up". There is an order that always works:
Solve these linear equations by expanding the brackets first. Remember to apply the distributive property carefully: a) $3(x - 2) = 12$ b) $2(x + 4) + 5 = 17$ c)…
Solve these linear equations by expanding the brackets first. Remember to apply the distributive property carefully:
- a) $3(x - 2) = 12$
- b) $2(x + 4) + 5 = 17$
- c) $10 - (x - 3) = 8$
Solve the following linear equation, making sure to simplify all terms and pay attention to sign changes in front of fractions:…
Solve the following linear equation, making sure to simplify all terms and pay attention to sign changes in front of fractions:
$\frac{2(x - 3)}{5} - \frac{x - 1}{2} = \frac{3x - 4}{10} + 1$
Solve the following linear equation with fractional coefficients and nested terms. Pay close attention to handling the negative signs in front of fractions:…
Solve the following linear equation with fractional coefficients and nested terms. Pay close attention to handling the negative signs in front of fractions:
$$\frac{2(x-5)}{3} - \frac{3x - \frac{x+1}{2}}{5} = \frac{x-1}{6} - \frac{7}{10}$$Is there always a solution?
Almost always, but not quite. When you simplify, this can happen:
| In the end you get… | Example | It means |
|---|---|---|
| $ax = b$ with $a \neq 0$ | $3x = 9$ | One solution: $x = 3$ |
| $0x = b$ with $b \neq 0$ | $0x = 5$ | No solution (no number times 0 gives 5) |
| $0x = 0$ | $0x = 0$ | Infinitely many solutions: it was an identity |
The drawing anticipates an idea you will see in functions: each side is a straight line, and the solution is where they cross.
Given the equation with an unknown coefficient $k$: $$\frac{k \cdot x}{3} - 4 = \frac{x + 2}{2}$$ Calculate the value that $k$ must have so that the solution to the…
Given the equation with an unknown coefficient $k$:
$$\frac{k \cdot x}{3} - 4 = \frac{x + 2}{2}$$Calculate the value that $k$ must have so that the solution to the equation is $x = 6$. Once $k$ is found, check that the equality holds by substituting both values into the original equation.
Rearranging formulae
The same rules work for making a letter the subject of a formula. From $v = \frac{d}{t}$ (speed) we get $d = v \cdot t$ and $t = \frac{d}{v}$.
In literal equations, we must be able to rearrange for one variable in terms of the others. Make $x$ the subject of the following equation, assuming that all parameters…
In literal equations, we must be able to rearrange for one variable in terms of the others. Make $x$ the subject of the following equation, assuming that all parameters $a, b, c$ and $d$ are known and non-zero, with $a \neq d$:
$$a(x + b) - c = d(x - a)$$Express the final result as a single fraction.
Find the length of side $x$ in the following rectangle, knowing that its total perimetre is $34$ cm.
Find the length of side $x$ in the following rectangle, knowing that its total perimetre is $34$ cm.
Back to the start: each mystery box weighs 3 kg, because if you take 2 kg off each pan you are left with 3 boxes weighing 9 kg ($3x = 9$), and sharing that between the 3 boxes gives $x = 3$ kg.