2 Equations, formulae and identities

Translate operations in their stated order. ‘Add 7 to $x$, then divide by 5’ is $(x+7)/5$, not $x+7/5$.

Substitution replaces a symbol by a value while preserving brackets: if $x=-2$, then $x^2=(-2)^2=4$.

Only like terms combine by addition or subtraction: $3x+5x=8x$, but $3x+5y$ cannot be simplified to $8xy$.

Multiplication signs are usually omitted between a number and letters; write $8ky$, with numerical coefficient first and letters in a consistent order.

Moving one step down in the index divides by the base: $x^3,x^2,x^1,x^0,x^{-1}$ gives $x^3,x^2,x,1,1/x$.

$2^{-3}=1/2^3=1/8$. A negative index creates a reciprocal; it does not make the value negative.

Handle numerical coefficients separately: $(6x^5)/(2x^2)=3x^3$.

A subtraction producing a negative index can be rewritten reciprocally: $x^2/x^5=x^{-3}=1/x^3$.

$(16x^8y^6)^{1/2}=16^{1/2}x^{8/2}y^{6/2}=4x^4y^3$ under the usual real-domain assumptions.

For real values, an even root requires a non-negative radicand; a negative power also requires a non-zero base.

If $a=-3$ and $b=4$, then $2a^2-b=2(-3)^2-4=18-4=14$. The brackets ensure that the square applies to the whole value $-3$.

Use the same value every time a symbol occurs. If $x=1/2$, then $4x^2+3x=4(1/2)^2+3(1/2)=5/2$.

$12x-7y-5x+2y=(12-5)x+(-7+2)y=7x-5y$.

Constants are also like terms with each other. Keep unlike groups separate and write the simplified expression in a clear order.

Draw or mentally track one multiplication for each term in the bracket, multiply coefficients and letter factors separately, then simplify.

The number of expanded terms should initially match the number inside the bracket; substituting a simple value can check equivalence.

$12x^3y-18x^2y^2=6x^2y(2x-3y)$. Expanding the result reproduces both original terms.

For $m^2+7m$, both terms contain $m$, so $m^2+7m=m(m+7)$.

Adding the four products gives $y^2-4y+9y-36=y^2+5y-36$.

$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$. This is a check on the leading, middle and constant terms, not a shortcut for omitting products.

For $x^2-5x-36$, the required numbers are $4$ and $-9$: their product is $-36$ and sum is $-5$. Hence $(x+4)(x-9)$.

Re-expand the brackets: the outer and inner products must combine to the original middle term.

Keep brackets around each intermediate polynomial and align like powers before collecting terms. A grid is useful when an intermediate factor has three or more terms.

The highest-degree term comes from multiplying all highest-degree terms; the constant term comes from multiplying all constants when none of the factors is just a variable.

$6x^2+11x+3$: since $ac=18$ and $9+2=11$, write $6x^2+9x+2x+3=3x(2x+3)+1(2x+3)=(3x+1)(2x+3)$.

$4c^2-9d^2=(2c)^2-(3d)^2=(2c-3d)(2c+3d)$.

$5/3-(x+2)/(2x)=[10x-3(x+2)]/(6x)=(7x-6)/(6x)$, where $x
e0$.

Cancellation is division by a common non-zero factor: $(x^2-9)/(x^2+3x)=[(x-3)(x+3)]/[x(x+3)]=(x-3)/x$, with $x
e0,-3$.

$3x^2-12x+7=3(x^2-4x)+7=3[(x-2)^2-4]+7=3(x-2)^2-5$.

From $a(x-h)^2+k$, the turning point is $(h,k)$ and the line of symmetry is $x=h$. If $a>0$ the minimum value is $k$; if $a<0$ the maximum is $k$.

Let $T_n=n(n+1)/2$. Then $T_n+T_{n+1}=n(n+1)/2+(n+1)(n+2)/2=(n+1)[n+n+2]/2=(n+1)^2$. Therefore the sum of two consecutive triangular numbers is a square.

Useful definitions include consecutive integers $n,n+1$, even integers $2n$, odd integers $2n+1$, and multiples of $k$ as $kn$.

Within one statement, repeated occurrences of the same letter represent the same value unless the letter is explicitly redefined.

The context may restrict possible values: a length is non-negative, a count is an integer, and a denominator cannot be zero.

Write numerical coefficients first and letter factors in a consistent order. Multiplication is implied by adjacency, but addition and subtraction remain explicit.

Use $=$ only between expressions known to have equal value. An expression such as $3x+2$ does not need an equals sign by itself.

If $T=5m-6n$, $m=4.2$ and $n=-2.5$, then $T=5(4.2)-6(-2.5)=21+15=36$.

For fractional values, keep exact fractions until the end when practical. Brackets also distinguish $(-3)^2$ from $-3^2$.

Potatoes cost 2 dollars per kg and carrots 3 dollars per kg. Buying $p$ kg and $c$ kg gives total cost $T=2p+3c$.

Define every symbol, match coefficients to their quantities, and test the formula with a simple numerical case and dimensional units.

Alisa picks $C$ cucumbers, Jena picks $C-5$, and Mikael picks $2C$. Therefore $T=C+(C-5)+2C=4C-5$.

For a perimeter, include every side before collecting; for area or volume, multiply the relevant dimensions. State the final subject explicitly, such as $T=\ldots$.

From $d=g+2ac$, subtract $g$ to get $d-g=2ac$, then divide by $2c$: $a=(d-g)/(2c)$, where $c
e0$.

Undo operations in reverse order. If necessary, clear a fraction first, but multiply every term on both sides consistently.

If $y=\sqrt{(x+1)/(x-4)}$, then $y^2(x-4)=x+1$. Collecting gives $x(y^2-1)=4y^2+1$, so $x=(4y^2+1)/(y^2-1)$, where the formula is defined.

For an even power, both roots may be needed unless the context restricts the subject, such as a positive length. A stated condition like $n>0$ selects the positive root.

Solve $(8-2x)/3-(2x-3)/2=4$. Multiply every term by 6: $2(8-2x)-3(2x-3)=24$. Then $25-10x=24$, so $x=1/10$.

Substitute the solution into both sides of the original equation, not only the simplified line. Equal results check signs, brackets and denominators.

A regular hexagon has side $(x-1)$ cm. An isosceles triangle has equal sides $(x+5)$ cm and base $(2x-3)$ cm. Equal perimeters give $6(x-1)=2(x+5)+(2x-3)$. Hence $x=6.5$, so each hexagon side is $5.5$ cm.

For triangle angles $a$, $a+10$ and $a+20$, use their total: $a+(a+10)+(a+20)=180$. Solve for $a$, then check all three angles are valid.

For $7x-2y=34$ and $3x+5y=-3$, multiply the first by 5 and the second by 2: $35x-10y=170$ and $6x+10y=-6$. Adding gives $41x=164$, so $x=4$ and then $y=-3$.

Keep fractions exact rather than rounding intermediate results. After finding one variable, substitute into an original equation to find the other.

For $(x+2y)/3=5$ and $2x-y/2=4$, first write $x+2y=15$ and $4x-y=8$. This exposes integer coefficients before elimination.

If elimination produces a false statement such as $0=5$, the lines are parallel and there is no solution. If it produces $0=0$, the equations describe the same line and have infinitely many solutions.

For $y-x-2=0$ and $2y+x=1$, the lines are $y=x+2$ and $y=(1-x)/2$. They intersect at $(-1,1)$, so $x=-1$ and $y=1$.

Intersecting lines give one solution; distinct parallel lines give none; coincident lines give infinitely many. A graph may give only an approximate coordinate unless the intersection is exactly readable.

For $x^2-5x-36=0$, use $4$ and $-9$: $(x+4)(x-9)=0$. Hence $x=-4$ or $x=9$.

The zero-product rule works because a product is zero only when at least one factor is zero. It applies after the equation has been set equal to zero.

Solve $6x^2-x-2=0$. Since $ac=-12$ and $3+(-4)=-1$, write $6x^2+3x-4x-2=0$, so $(3x-2)(2x+1)=0$. Thus $x=2/3$ or $x=-1/2$.

Expand the factors before applying the zero-product rule. The leading, middle and constant coefficients must exactly match the standardised equation.

For $2x^2+3x-1=0$, $a=2,b=3,c=-1$, so $x=[-3\pm\sqrt{17}]/4$. Keep the entire numerator over $2a$.

For $x^2-6x+5=0$, write $(x-3)^2-4=0$. Then $(x-3)^2=4$, so $x=3\pm2$, giving $1$ and $5$.

If $D>0$ there are two real roots; if $D=0$ one repeated real root; if $D<0$ no real roots. An exact surd should not be rounded unless requested.

A trapezium has parallel sides $x+5$ and $3x-2$, height $2x-3$, and area 133. Then $[(x+5)+(3x-2)](2x-3)/2=133$, which simplifies to $8x^2-6x-275=0$.

Solve the derived equation, then return to every original expression. Reject a root that makes a length non-positive, violates a denominator restriction, or conflicts with the stated domain.

For $y=3-2x$ and $x^2+y^2=18$, substitution gives $x^2+(3-2x)^2=18$, hence $5x^2-12x-9=0=(5x+3)(x-3)$. The solutions are $(-0.6,4.2)$ and $(3,-3)$.

The solutions are intersection points of a line and a conic. Two roots mean two intersections, a repeated root means tangency, and no real roots means no real intersection.

If only integer values are requested, list integers satisfying both ends. For $-3.4<n\le2$, the values are $-3,-2,-1,0,1,2$.

Equivalent statements can reverse order and symbol together: $x>3$ means $3<x$. Read each relation from left to right to check meaning.

To represent $-2<y\le5$, mark an open endpoint at $-2$, a closed endpoint at $5$, and join the interval between them.

When reading a diagram, identify each boundary value, whether its marker is open or closed, and whether the set extends left, right or between endpoints before writing symbols.

Solve $7-3t<2t+15$: $-5t<8$. Dividing by $-5$ reverses the sign, so $t>-8/5$.

For $-5\le2p+3<13$, subtract 3 from every part and divide every part by 2: $-4\le p<5$.

After solving, use an open or closed endpoint and shade the correct direction or interval. Test a simple value if the direction is uncertain.

For $x\le6$, draw the vertical line $x=6$ and select its left side. For $y\ge2$, draw the horizontal line $y=2$ and select above it.

For $y\le x+1$, the boundary is $y=x+1$. Since $(0,0)$ satisfies $0\le1$, shade the side containing the origin.

A region bounded by $x=-1$, $x+y=4$ and $y=x/3-2$ that lies right of the vertical, below the descending line and above the rising line is $x\ge-1$, $x+y\le4$, $y\ge x/3-2$.

Every point in the region must satisfy all inequalities simultaneously. Check one interior point against the complete list and verify that each boundary contributes an edge.

For this Foundation objective, syllabus conventions for inclusion of Cartesian boundaries are not required; the essential work is selecting the correct side of every line.

For $5y^2-17y\le40$, factor $5y^2-17y-40=(5y+8)(y-5)$. The upward-opening quadratic is non-positive between its roots, so $-8/5\le y\le5$.

For an upward-opening quadratic, $f(x)>0$ usually lies outside two distinct roots and $f(x)<0$ between them; verify rather than memorise if the leading coefficient is negative.

In a contextual problem, intersect the algebraic solution with restrictions such as positive lengths or a non-zero denominator.

For $x\le4$, $y\le2x+1$ and $5x+2y\le20$, draw $x=4$, $y=2x+1$ and $5x+2y=20$, test the correct side of each, then keep only their common overlap.

Candidate vertices come from pairwise boundary intersections. A visible intersection is part of the feasible region only if it satisfies all remaining inequalities.

A feasible region may be unbounded or empty. Do not force a closed polygon when the half-planes do not create one.