The instructional goals for this section identify three key skills. First, learners must determine whether a given ordered pair acts as a valid solution for a system of linear inequalities. Second, they will actively construct solutions by graphing systems of linear inequalities. Finally, they will apply these skills to solve real-world applications structured around systems of inequalities.

Learning Objectives for Graphing Systems of Linear Inequalities

To solve the linear inequality $$2a < 5a + 12$$, you must isolate the variable $$a$$. Start by subtracting $$5a$$ from both sides to gather the variable terms on the left side of the inequality, resulting in $$-3a < 12$$. Then, divide both sides by $$-3$$. Remember that when dividing or multiplying an inequality by a negative number, you must reverse the inequality symbol. Thus, the sign changes from $$<$$ to $$>$$, giving the final solution of $$a > -4$$.

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Most linear inequalities require more than one step to solve. The process follows the same general strategy used for solving linear equations: apply properties of inequality to simplify both sides, and securely collect all variable terms on one side and all constant terms on the other. A critical difference arises when isolating the variable at the end: if both sides must be multiplied or divided by a negative number to solve, the direction of the inequality sign must be reversed.

Solving Multi-Step Linear Inequalities

Before delving into graphing linear inequalities in two variables, an instructional readiness quiz checks the student's prerequisite ability to solve single-variable inequalities. The student is prompted to solve the inequality $$4x + 3 > 23$$. This standard exercise requires the student to properly isolate the variable by first subtracting $$3$$ from both sides to obtain $$4x > 20$$, and subsequently dividing by $$4$$ to securely discover the solution $$x > 5$$. This step actively tests whether the learner possesses the fundamental algebraic skills needed to proceed to more advanced graphing.

Readiness Quiz: Solving $$4x + 3 > 23$$

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OpenStax Intermediate Algebra

To solve the inequality $$8p + 3(p - 12) > 7p - 28$$, first simplify each side by distributing to get $$8p + 3p - 36 > 7p - 28$$, which condenses to $$11p - 36 > 7p - 28$$. Next, collect the variables on the left side by subtracting $$7p$$ from both sides to obtain $$4p - 36 > -28$$. Add $$36$$ to both sides to collect the constants on the right, yielding $$4p > 8$$. Divide both sides by $$4$$. Since $$4$$ is notably positive, the inequality symbol remains exactly the same, culminating in $$p > 2$$. Graph the solution on the number line with a left parenthesis at $$2$$ and shading stretching to the right. In interval notation, the solution is written securely as $$(2, \infty)$$.

Example: Solving and Graphing $$8p + 3(p - 12) > 7p - 28$$

To solve the multi-step linear inequality $$9y + 2(y + 6) > 5y - 24$$, start by distributing to simplify the left side, which generates $$9y + 2y + 12 > 5y - 24$$ and combines smoothly to $$11y + 12 > 5y - 24$$. Subtract $$5y$$ from both sides to responsibly collect the variables on the left, leading directly to $$6y + 12 > -24$$. By subtracting $$12$$ from both sides, the constants are gathered firmly on the right, resulting in $$6y > -36$$. Finally, divide both sides by $$6$$ to comprehensively isolate the variable. Because $$6$$ is positive, the inequality's direction is seamlessly preserved, outputting $$y > -6$$. The solution is then efficiently graphed on a number line using a left parenthesis at $$-6$$ with shading strictly to the right. Under interval notation, the solution is explicitly captured as $$(-6, \infty)$$.

Example: Solving and Graphing $$9y + 2(y + 6) > 5y - 24$$

To solve the linear inequality $$6u + 8(u - 1) > 10u + 32$$, first aggressively apply distribution to clarify the left side into $$6u + 8u - 8 > 10u + 32$$. Combine associated like terms to confidently gather $$14u - 8 > 10u + 32$$. To draw the variables collectively onto the left, logically subtract $$10u$$ from both sides, producing $$4u - 8 > 32$$. Add $$8$$ to both sides to purposefully reposition the constant onto the right, obtaining $$4u > 40$$. Divided by $$4$$, the variable is neatly isolated. Since $$4$$ is positive, the inequality symbol acts unchanged, concluding firmly with $$u > 10$$. This exact solution is graphed featuring a left parenthesis reliably at $$10$$ followed by distinct shading constantly to the right. In interval notation, the entire outcome is formally noted as $$(10, \infty)$$.

Example: Solving and Graphing $$6u + 8(u - 1) > 10u + 32$$

To solve the inequality $$8x - 2(5 - x) < 4(x + 9) + 6x$$, begin faithfully by simplifying each side. Proper distribution rapidly produces $$8x - 10 + 2x < 4x + 36 + 6x$$. Consolidating like terms correctly cuts this parameter down to $$10x - 10 < 10x + 36$$. Subtracting $$10x$$ from both sides abruptly eliminates the variable entirely, producing the final numerical statement $$-10 < 36$$. Because this represents a mathematically true statement unconditionally, the inequality formally operates as an identity, signifying that its valid solution strictly comprises all real numbers. On a typical number line, this graphically involves shading the line altogether. In functional interval notation, this definitive solution is visibly transcribed as $$(-\infty, \infty)$$.

Example: Solving and Graphing $$8x - 2(5 - x) < 4(x + 9) + 6x$$

To solve the linear inequality $$4b - 3(3 - b) > 5(b - 6) + 2b$$, systematically simplify both sides heavily by distributing, extracting $$4b - 9 + 3b > 5b - 30 + 2b$$. Combining corresponding like terms accurately returns $$7b - 9 > 7b - 30$$. Subtracting $$7b$$ logically from both sides completely zeroes out the variable terms and leaves only $$-9 > -30$$. Since this denotes an inherently true mathematical statement irrespective of variables, the unique inequality robustly functions as an identity boasting an infinite solution of all real numbers. The entire number line is permanently shaded for accurate geometric representation, natively defining $$(-\infty, \infty)$$ in standard interval notation.

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