Honors Pre-Calculus Unit 9 Review: Systems of Equations and Inequalities

A system of linear equations is two or more equations that share the same variables. A solution is an ordered pair that makes every equation true at once. On a graph, each equation is a line, so the solution is where the lines cross.

You have three main ways to solve. Graphing shows the picture but can be imprecise. Substitution works best when a variable already has a coefficient of 1 or -1. Elimination works best when both equations are in standard form $Ax + By = C$.

Worked example. Solve $2x + 3y = 12$ and $x - y = 1$. Multiply the second equation by 3 to get $3x - 3y = 3$. Add it to the first equation and the $y$ terms cancel, leaving $5x = 15$, so $x = 3$. Substitute into $x - y = 1$ to get $y = 2$. Check in the first equation, where $2(3) + 3(2) = 12$. The solution is $(3, 2)$.

The algebra also tells you what type of system you have. A false statement like $0 = 5$ means the lines are parallel and there's no solution. A true statement like $0 = 0$ means both equations describe the same line, so every point on it works. You can write those infinitely many solutions with a parameter, such as $(t,\ 2t - 3)$ for any real $t$.

With three variables, each linear equation describes a plane in 3D space, and the solution is where all three planes meet. The strategy is to turn the three-variable system into a two-variable system you already know how to solve.

Pick one variable to eliminate. Use two different pairs of equations to cancel it, which gives you two new equations in the remaining two variables. Solve that smaller system, then substitute back to find the third variable.

Worked example. Solve $x + y + z = 6$, $2x - y + z = 3$ and $x + 2y - z = 2$. Add the first and third equations to cancel $z$, giving $2x + 3y = 8$. Add the second and third to cancel $z$ again, giving $3x + y = 5$. From the second new equation, $y = 5 - 3x$. Substitute to get $2x + 15 - 9x = 8$, so $x = 1$ and $y = 2$. Then $1 + 2 + z = 6$ gives $z = 3$. The solution is $(1, 2, 3)$.

The same three outcomes from two variables still apply. If one equation is a combination of the others and the system is consistent, the solutions form a line (one parameter) or a plane (two parameters). A contradiction like $0 = 5$ means no point lies on all three planes, even if an equation is redundant.

A nonlinear system has at least one equation that is not linear, such as $x^2 + y^2 = 25$. Solutions are still intersection points, but there can be none, a finite number, or infinitely many. Substitution is often useful.

Worked example. Solve $x^2 + y^2 = 25$ and $y = x + 1$. Substitute to get $x^2 + (x + 1)^2 = 25$, which expands to $2x^2 + 2x - 24 = 0$, or $x^2 + x - 12 = 0$. Factor to $(x + 4)(x - 3) = 0$, so $x = 3$ or $x = -4$. Use $y = x + 1$ to get the points $(3, 4)$ and $(-4, -3)$. Both check in the circle equation, since $9 + 16 = 25$ and $16 + 9 = 25$.

To graph a system of inequalities, graph each boundary first. Use a solid curve for $\le$ or $\ge$ and a dashed curve for $<$ or $>$. Test a point in each region separated by the boundary and shade the regions where the inequality is true. The solution is where all the shadings overlap.

For example, $x^2 + y^2 < 9$ is the inside of a circle of radius 3, and $y \ge x^2$ is the region on and above the parabola. Their overlap is the part of the circle's interior above the parabola.

Partial fraction decomposition runs fraction addition in reverse. You take one rational expression and split it into a sum of simpler fractions. You'll use this in calculus when integrating rational functions.

First make sure the fraction is proper, meaning the numerator's degree is less than the denominator's. If it isn't, do polynomial long division first. Keep the polynomial quotient and decompose the remainder divided by the original denominator. Then factor the denominator completely and choose the form.

Each distinct linear factor $(x - a)$ gets a term $\frac{A}{x - a}$. A repeated factor $(x - a)^n$ needs one term for every power from 1 to $n$. An irreducible quadratic factor like $x^2 + 4$ gets a linear numerator, $\frac{Bx + C}{x^2 + 4}$. If that quadratic is repeated, include a term with a linear numerator for each power of the factor.

Worked example. Decompose $\frac{5x + 1}{(x - 1)(x + 2)}$. Set it equal to $\frac{A}{x - 1} + \frac{B}{x + 2}$ and multiply both sides by $(x - 1)(x + 2)$ to get $5x + 1 = A(x + 2) + B(x - 1)$. Let $x = 1$, so $6 = 3A$ and $A = 2$. Let $x = -2$, so $-9 = -3B$ and $B = 3$. The answer is $\frac{2}{x - 1} + \frac{3}{x + 2}$.

When plugging in values doesn't isolate every constant, expand and match coefficients of each power of $x$. That creates a system of equations, which is why this topic lives in this unit.

A matrix is a rectangular grid of numbers, described by its dimensions as rows by columns. A $2 \times 3$ matrix has 2 rows and 3 columns.

You can add or subtract matrices only when they have the same dimensions, and you do it entry by entry. Scalar multiplication multiplies every entry by the same number.

Multiplication follows a different rule. The product $MN$ exists only when the number of columns in $M$ equals the number of rows in $N$. A $2 \times 3$ matrix times a $3 \times 4$ matrix gives a $2 \times 4$ matrix. Each entry of the product comes from a row of $M$ and a column of $N$, multiplied entry by entry and added.

Worked example. Let $M = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $N = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$. The top-left entry of $MN$ is $1(5) + 2(7) = 19$. Finishing the rest gives $MN = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$. Computing $NM$ gives $\begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix}$, so $MN \neq NM$.

An identity matrix has 1s on the main diagonal and 0s elsewhere. Multiplying $A$ by an identity matrix of a compatible size on either side leaves $A$ unchanged. Any linear system can be written as $AX = B$, where $A$ holds the coefficients, $X$ the variables and $B$ the constants.

An augmented matrix holds a whole system in one grid. Each row is an equation, each column before the bar is a variable, and the last column holds the constants. If a variable is missing from an equation, its entry is 0.

Three row operations change the matrix without changing the solutions. You can swap two rows, multiply a row by a nonzero constant, or add a multiple of one row to another. Gaussian elimination uses these to reach row echelon form, a staircase shape with zeros below each leading entry. Then you back-substitute from the bottom row up. Gauss-Jordan elimination keeps going to reduced row echelon form, with zeros above and below each leading 1, so you can read the answer directly.

Worked example. Solve $x + y = 5$ and $2x - y = 4$. The augmented matrix has rows $(1, 1 \mid 5)$ and $(2, -1 \mid 4)$. Replace row 2 with row 2 minus 2 times row 1, which gives $(0, -3 \mid -6)$. That row says $-3y = -6$, so $y = 2$. Row 1 says $x + 2 = 5$, so $x = 3$.

The final matrix also classifies the system. A row like $(0, 0, 0 \mid 5)$ means $0 = 5$, so there's no solution. In row echelon form, a variable column with no leading entry means that variable is free. If the system is consistent and has a free variable, it has infinitely many solutions.

The inverse of a square matrix $A$, written $A^{-1}$, satisfies $AA^{-1} = A^{-1}A = I$. If you have $AX = B$, multiply both sides on the left by $A^{-1}$ to get $X = A^{-1}B$.

For a $2 \times 2$ matrix with rows $(a, b)$ and $(c, d)$, the determinant is $ad - bc$. If it's not zero, swap $a$ and $d$, change the signs of $b$ and $c$, and divide every entry by the determinant. If the determinant is zero, the matrix is singular and has no inverse. For a $3 \times 3$ matrix, write $[A \mid I]$ and row reduce until the left side becomes $I$. The right side is then $A^{-1}$.

Worked example. Solve $3x + y = 7$ and $2x + 4y = 8$. The coefficient matrix has rows $(3, 1)$ and $(2, 4)$, with determinant $3(4) - 1(2) = 10$. So $A^{-1}$ has rows $\frac{1}{10}(4, -1)$ and $\frac{1}{10}(-2, 3)$. Multiply by the constants $(7, 8)$. Then $x = \frac{1}{10}(28 - 8) = 2$ and $y = \frac{1}{10}(-14 + 24) = 1$. Check with $3(2) + 1 = 7$.

This method pays off when several systems share the same coefficients but have different constants. You find $A^{-1}$ once and reuse it.

Cramer's Rule solves a square system using only determinants. Let $D$ be the determinant of the coefficient matrix. To find a variable, replace that variable's column with the constants, take the determinant, and divide by $D$. So $x = \frac{D_x}{D}$, $y = \frac{D_y}{D}$, and for three variables $z = \frac{D_z}{D}$.

For a $3 \times 3$ matrix with rows $(a, b, c)$, $(d, e, f)$ and $(g, h, i)$, expand along the first row to get $a(ei - fh) - b(di - fg) + c(dh - eg)$. Keep the alternating $+, -, +$ signs.

Worked example. Solve $2x + 3y = 12$ and $x - y = 1$. Then $D = 2(-1) - 3(1) = -5$. Replace the $x$ column with $(12, 1)$, so $D_x = 12(-1) - 3(1) = -15$ and $x = \frac{-15}{-5} = 3$. Replace the $y$ column instead, so $D_y = 2(1) - 12(1) = -10$ and $y = \frac{-10}{-5} = 2$. That matches the elimination answer from 9.1.

If $D = 0$, Cramer's Rule doesn't apply, because the system has no solution or infinitely many. Use row reduction to find out which one.