Russian state exam, grade 9

Equations, inequalities, calculations and part 2: systems, fractions, factoring.

Here are the algebra tasks: No. 8, 9, 13 and the algebra task No. 20 of part 2. Geometry and word problems are not here yet.

No. 8 Calculations with powers OpenClose

Find the value of an expression with powers. Bring the powers to one base, and everything cancels.

How to solve it

Expressions with powers

An expression like $\frac{2^5 \cdot 4^3}{8^3}$ looks heavy, but once every number is written as a power of two it folds into a single power. No big multiplications needed.

Step by step

Bring all the powers to one base: $4 = 2^2$, $8 = 2^3$, $\frac{1}{2} = 2^{-1}$.
When multiplying powers add the exponents, when dividing subtract them, when raising a power to a power multiply them.
Work out the final exponent and compute the power.

The exponent of the first power.

The exponent of the second power. The rules hold only when the bases are equal.

Example: $\frac{2^5 \cdot 4^3}{8^3} = \frac{2^5 \cdot 2^6}{2^9} = 2^{5 + 6 - 9} = 2^2 = 4$.

Common mistakes

Multiplying the exponents when multiplying powers: $2^3 \cdot 2^4 = 2^7$, not $2^{12}$.
Merging powers with different bases: $2^3 \cdot 3^2$ is not a single power.
Confusing a negative exponent with a negative number: $2^{-3} = \frac{1}{8}$, not $-8$.

Example

No. 9 Equations OpenClose

A linear, quadratic or fractional equation.

How to solve it

Linear equations

In a linear equation the unknown appears only to the first power: $3x + 5 = x - 7$. It has one root, and a few moves find it.

Step by step

Open any brackets.
Move the $x$ terms to the left and the numbers to the right. A term changes its sign as it crosses the «=»: $3x + 5 = x - 7$ becomes $3x - x = -7 - 5$.
Collect like terms: $2x = -12$.
Divide both sides by the number in front of $x$: $x = -6$.
Check by substituting: the left side is $3 \cdot (-6) + 5 = -13$, the right side is $-6 - 7 = -13$. They match.

The number in front of $x$, its coefficient. You may divide by it only when $a \ne 0$.

Whatever is left on the right after moving and collecting terms.

If $x$ cancels out and you get a false statement like $0 = 5$, there are no roots. If you get $0 = 0$, every number is a root.

Common mistakes

Moving a term to the other side and keeping its sign.
Losing a minus when dividing: $-2x = 8$ gives $x = -4$, not $4$.
Opening a bracket with a minus in front and changing only the first sign. Correct: $-(x - 3) = -x + 3$.

Example

Quadratic equations

A quadratic equation looks like $ax^2 + bx + c = 0$ with $a \ne 0$. It has at most two roots, and one number, the discriminant, tells you how many there really are.

Step by step

Move everything to the left so that the right side is zero. Write down $a$, $b$ and $c$ with their signs.
If $c = 0$, take $x$ out of the brackets: $x(ax + b) = 0$, so the roots are $0$ and $-\frac{b}{a}$. If $b = 0$, solve for $x^2$ and take the square root.
Otherwise compute the discriminant $D = b^2 - 4ac$.
If $D > 0$ there are two roots, if $D = 0$ one, if $D < 0$ no real roots.
Check with Vieta's formulas: the roots add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$.

The coefficient of $x$ with its sign. In $x^2 - 5x + 6 = 0$ it is $b = -5$, so $-b = 5$.

The discriminant: it decides how many roots there are.

The coefficient of $x^2$.

Example: $x^2 - 5x + 6 = 0$, $D = 25 - 24 = 1$, $x = \frac{5 \pm 1}{2}$, so $x = 2$ and $x = 3$.

Common mistakes

Dropping the sign of $b$: in $x^2 - 5x + 6$ the coefficient is $b = -5$, and $b^2 = 25$.
Computing the discriminant before the right side is zero. Move everything to one side first.
Dividing both sides by $x$ and losing the root $x = 0$. Take $x$ out of the brackets instead.

Example

Equations with fractions

In an equation with fractions the unknown sits in a denominator: $\frac{x^2 - 1}{x - 1} = 0$. You cannot divide by zero, and the whole solution starts from that rule.

Step by step

Find the values of $x$ that make a denominator zero. They are forbidden; the rest of the numbers form the domain of the equation.
Move everything to one side and bring it to a common denominator, so that you have a single fraction.
A fraction is zero when its numerator is zero. Solve «numerator $= 0$».
Throw away the roots that were forbidden in the first step.

The numerator: set it equal to zero.

The denominator: it must not be zero, so such roots are dropped.

Example: in $\frac{x^2 - 1}{x - 1} = 0$ the numerator gives $x = 1$ and $x = -1$, but $x = 1$ makes the denominator zero. Answer: $x = -1$.

Common mistakes

Keeping a root that makes a denominator zero. This is the most common mistake.
Cancelling an expression with $x$ in it and forgetting that it could not be zero.
Multiplying only some of the terms by the missing factor when finding the common denominator.

Example

No. 13 Inequalities OpenClose

A linear or quadratic inequality. The answer is an interval.

How to solve it

Linear inequalities

A linear inequality is solved almost like a linear equation: $3x - 5 > x + 1$. There is one difference, and it matters: when you multiply or divide by a negative number, the inequality sign flips.

Step by step

Move the $x$ terms to the left and the numbers to the right, changing the signs of the moved terms as in an equation.
Collect like terms to get $ax > b$ (or another inequality sign).
Divide both sides by $a$. If $a > 0$, the sign stays. If $a < 0$, it flips.
Write the answer as an interval: $x > 3$ is $(3, +\infty)$, and $x \le 3$ is $(-\infty, 3]$.

The number in front of $x$. Its sign decides whether the inequality flips. Why: multiplying $2 < 3$ by $-1$ gives $-2 > -3$.

Example: $-2x > 6$. Divide by $-2$ and flip the sign: $x < -3$. Answer: $(-\infty, -3)$.

Common mistakes

Dividing by a negative number without flipping the sign.
Mixing up brackets: with a strict sign ($>$ or $<$) the end is not included and the bracket is round; with $\ge$ or $\le$ it is included and the bracket is square.
Putting a square bracket next to infinity. Infinity always gets a round bracket.

Example

Quadratic inequalities

A quadratic inequality has a quadratic on the left: $x^2 - 5x + 6 \le 0$. Its graph is a parabola, and the answer is easiest to read straight off the picture: where the parabola is above the axis and where it is below.

Step by step

Move everything to the left so that the right side is zero.
Find the roots of the quadratic by solving $ax^2 + bx + c = 0$. At the roots the parabola crosses the axis.
Sketch the parabola: it opens upwards when $a > 0$ and downwards when $a < 0$.
For $>$ or $\ge$ take the parts where the parabola is above the axis; for $<$ or $\le$, where it is below.
Include the roots only when the inequality is not strict.

The leading coefficient. When it is positive, the parabola opens upwards: it is below the axis between the roots and above it outside. For $a < 0$ it is the other way round.

The roots of the quadratic, $x_1 < x_2$.

Example: $x^2 - 5x + 6 \le 0$. The roots are $2$ and $3$, the parabola opens upwards, and we need the points below the axis together with the roots: $[2, 3]$.

Common mistakes

Forgetting that a negative $a$ turns the parabola upside down, and taking the wrong part.
Mixing up «between the roots» and «outside». A quick check helps: substitute one number from your answer.
Not knowing what to do when $D < 0$. Then the quadratic has the same sign everywhere, and the answer is either every number or the empty set.

Example

No. 20 Systems of equations OpenClose

A system of two equations in two unknowns. The answer is every pair (x, y).

How to solve it

Systems of linear equations

A system of two linear equations, such as $x + 2y = 7$ and $3x - y = 7$, puts two conditions on the same pair of numbers $(x, y)$. You need the pair that satisfies both. The usual tools are substitution and adding the equations.

Step by step

Find an equation where one variable is easy to express (coefficient $1$ or $-1$) and express it: $x + 2y = 7$ gives $x = 7 - 2y$.
Substitute it into the other equation: $3(7 - 2y) - y = 7$. One unknown is left.
Solve it: $21 - 7y = 7$, $y = 2$. Then find the other variable: $x = 7 - 4 = 3$.
If expressing is awkward, add the equations after multiplying them so that the coefficients of one variable become opposite.
Check the pair in both equations and write the answer: $(3, 2)$.

The coefficients of $x$ are opposite, so adding the equations removes $x$.

Then $y = 2$, and either equation gives $x = \frac{1}{2}$. Check: $2 \cdot \frac{1}{2} + 3 \cdot 2 = 7$.

Common mistakes

Multiplying only one side of an equation by a number; both sides must be multiplied.
Substituting the found number and not checking the pair in the other equation.
Writing the pair in the wrong order: $x$ comes first, then $y$.

Example

Nonlinear systems

In a nonlinear system one equation is linear and the other is not: $x + y = 1$, $xy = -12$. There are usually two solutions, each a pair of numbers. Substitution works: express a variable from the linear equation and put it into the other.

Step by step

Express one variable from the linear equation: $y = 1 - x$.
Substitute it into the other equation: $x(1 - x) = -12$.
Solve the quadratic equation you get: $x^2 - x - 12 = 0$, with roots $4$ and $-3$.
For each root find the other variable: $y = 1 - 4 = -3$ and $y = 1 - (-3) = 4$. That gives two pairs.
Check both pairs and write them in the answer: $(4, -3)$ and $(-3, 4)$.

The sum of the unknowns.

Their product. This is Vieta's theorem read backwards.

Example: $x + y = 1$, $xy = -12$ give $t^2 - t - 12 = 0$ with roots $4$ and $-3$.

Common mistakes

Finding only $x$ and forgetting to compute $y$.
Writing one pair, although the quadratic has two roots and so the system has two solutions.
Swapping $x$ and $y$ in a pair.

Example

No. 20 Fractions and factoring OpenClose

Reduce a fraction or factor a polynomial.

How to solve it

Reducing fractions

To reduce a fraction like $\frac{x^2 + 7x + 6}{x^2 + 5x + 4}$ is to divide its numerator and denominator by their common factor. Only factors can be cancelled, so both polynomials are factored first.

Step by step

Factor the numerator: $x^2 + 7x + 6 = (x + 1)(x + 6)$.
Factor the denominator: $x^2 + 5x + 4 = (x + 1)(x + 4)$.
Cancel the common factor: you get $\frac{x + 6}{x + 4}$.
Remember where the original fraction is undefined: the reduced one equals it everywhere except at these points (here $x \ne -1$ and $x \ne -4$).

The common factor of the numerator and the denominator. Both are divided by it.

The condition $x \ne -1$ stays from the original fraction: at $x = -1$ it is undefined, while the reduced one would be defined.

Common mistakes

Cancelling terms instead of factors: nothing cancels in $\frac{x + 6}{x + 4}$.
Not factoring completely and missing the common factor.
Missing factors that differ only in sign: $1 - x = -(x - 1)$.

Example

Factoring

To factor a polynomial is to write it as a product of simpler ones: $x^2 - 5x + 6 = (x - 2)(x - 3)$. You need it on its own, to reduce fractions and to solve equations.

Step by step

Take out a common factor if there is one: $2x^2 - 6x = 2x(x - 3)$.
Look for the special products, for instance $a^2 - b^2 = (a - b)(a + b)$.
A quadratic factors through its roots.
For a cubic, find an integer root among the divisors of the constant term, check it by substituting, and divide the polynomial by $(x - x_0)$.
Go on until every factor is linear or cannot be factored further.

The leading coefficient. It is easy to forget.

The roots of $ax^2 + bx + c = 0$.

Example: the roots of $x^2 - 5x + 6 = 0$ are $2$ and $3$, so $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Common mistakes

Losing the leading coefficient: $2x^2 - 10x + 12 = 2(x - 2)(x - 3)$, and without the $2$ the equality is false.
Getting the signs wrong: the root $2$ gives the factor $(x - 2)$, not $(x + 2)$.
Stopping halfway, when one of the factors still splits.