Algebra Study Guide

Inequalities extend the concept of equations by comparing values using symbols like <, >, ≤, or ≥. Solving inequalities in one variable means finding the range of values for that variable that make the inequality true. The core principles for solving inequalities are similar to solving equations: isolate the variable by performing inverse operations. However, a critical difference arises when multiplying or dividing both sides of an inequality by a negative number: the inequality sign must be reversed.

When dealing with linear inequalities, you typically aim to collect all terms involving the variable on one side and constant terms on the other, then simplify. For example, to solve $3x - 5 < x + 7$, you would subtract $x$ from both sides to get $2x - 5 < 7$, then add $5$ to both sides to get $2x < 12$, and finally divide by $2$ to get $x < 6$.

Quadratic inequalities, like $x^2 + 2x - 3 &gt; 0$, require a different approach. First, rearrange the inequality so that one side is zero. Then, find the roots of the corresponding quadratic equation ($x^2 + 2x - 3 = 0$). For this example, factoring gives $(x+3)(x-1) = 0$, so the roots are $x = -3$ and $x = 1$. These roots divide the number line into intervals. You then test a value from each interval in the original inequality to determine which intervals satisfy it. Alternatively, sketching the parabola $y = x^2 + 2x - 3$ shows where its graph is above or below the x-axis, directly indicating the solution set. For $x^2 + 2x - 3 > 0$, the parabola opens upwards, so the graph is above the x-axis for $x < -3$ or $x > 1$.

Compound inequalities involve two or more inequalities linked by 'and' or 'or'. For 'and' inequalities, such as $-2 ext{ extless } x ext{ extless } 5$, the solution must satisfy both parts simultaneously. For 'or' inequalities, such as $x ext{ extless } -2$ or $x ext{ extgreater } 5$, the solution satisfies at least one of the parts. Inequalities involving algebraic fractions often require careful consideration of denominators. It's usually best to avoid cross-multiplication if the sign of the denominator is unknown, as this can lead to incorrect sign reversals. Instead, rearrange to have zero on one side, combine the fractions, and then analyse the signs of the numerator and denominator.

Representing solutions on a number line is crucial: use an open circle for strict inequalities (&lt;, &gt;) and a closed circle for non-strict inequalities ($ ext{ extle}, ext{ extge}$). Arrows indicate the direction of the solution set.

Inequalities involving two variables, typically 'x' and 'y', define a specific region on a coordinate plane rather than a single line or point. Solutions to such inequalities are represented by shaded regions. When you have one inequality, say ax + by < c, the first step is to treat it as an equation ax + by = c and draw the corresponding line. This line acts as a boundary. If the inequality is strict (using < or >), the line should be drawn as a dashed or dotted line to indicate that points on the line itself are not included in the solution set. If the inequality is non-strict (using ≤ or ≥), the line should be drawn as a solid line, meaning points on the line are part of the solution. After drawing the boundary line, you need to determine which side of the line satisfies the inequality. A common method is to pick a test point not on the line, often the origin (0,0) if it isn't on the boundary line, and substitute its coordinates into the original inequality. If the inequality holds true for the test point, then the region containing that point is the solution region and should be shaded. If it does not hold true, then the region on the opposite side of the line is the solution region. When dealing with a system of two or more inequalities in two variables, each inequality defines its own region. The solution to the system is the region where all individual inequalities are simultaneously satisfied. This is the overlapping region, which should be clearly indicated, usually by shading. For example, if you have y > x and y < -x + 4, you would draw y=x (dashed) and y=-x+4 (dashed). Then test a point for each. For y > x, (0,1) gives 1 > 0 (True), so shade above y=x. For y < -x+4, (0,0) gives 0 < 4 (True), so shade below y=-x+4. The solution is the region where these two shaded areas overlap. Understanding the type of line (solid/dashed) and the correct shading direction is crucial for accurately representing the solution set of inequalities in two variables.