Algebra 1 — CSEC Maths notes

Substitution

Replace each letter with its value, then follow the order of operations.

If $x = 3$ , then $2x^2 - 5 = 2 \times 9 - 5 = 13$ .

Negative values need brackets when you substitute: for $x = -2$ , write $2(-2)^2$ , which is $8$ — not $-8$ .

Terms are like only if they have exactly the same letters to the same powers.

$5a + 3b - 2a + 7b = 3a + 10b$

Note that $3a$ , $3a^2$ and $3ab$ are all unlike one another.

Multiply everything inside the bracket by whatever is outside, watching the signs.

$3(2x - 1) - 2(x + 4) = 6x - 3 - 2x - 8 = 4x - 11$

For two brackets, every term in the first multiplies every term in the second — four products:

$(2x + 3)(x - 5) = 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15$

Factorising is expanding in reverse: put the brackets back.

Common factor — take out the largest thing every term shares. $6x^2 + 9x = 3x(2x + 3)$

Difference of two squares — two squared terms with a minus between them.

Do the same thing to both sides until the letter stands alone. Gather the letters on one side and the numbers on the other.

$5x - 7 = 2x + 8 \;\Rightarrow\; 3x - 7 = 8 \;\Rightarrow\; 3x = 15 \;\Rightarrow\; x = 5$

Same rules as solving an equation, but the answer is a formula rather than a number. Undo the operations in reverse order.

To make $t$ the subject of $v = u + at$ : subtract $u$ , then divide by $a$ .

$t = \frac{v - u}{a}$

To make $r$ the subject of $A = \pi r^2$ : divide by $\pi$ , then take the square root.

$r = \sqrt{\frac{A}{\pi}}$