Algebra 1 (HS pathway)

Linear Equations and Inequalities: Solving, Graphing, and Systems

Algebra 1

✓By the end of this lesson students will be able to solve linear equations in one variable.
✓By the end of this lesson students will be able to solve linear inequalities in one variable and graph their solutions on a number line.
✓By the end of this lesson students will be able to graph linear equations and inequalities in two variables on a coordinate plane.
✓By the end of this lesson students will be able to solve systems of linear equations using graphing, substitution, and elimination methods.
✓By the end of this lesson students will be able to solve systems of linear inequalities by graphing their solution regions.

Key concepts

Linear Equation in One Variable

An equation that can be written in the form ax + b = c, where a, b, and c are real numbers and a ≠ 0. The solution is a single value for the variable that makes the equation true.

ax + b = c

Linear Inequality in One Variable

An inequality that can be written in the form ax + b < c, ax + b > c, ax + b ≤ c, or ax + b ≥ c, where a, b, and c are real numbers and a ≠ 0. The solution is a range of values for the variable. When multiplying or dividing both sides by a negative number, the inequality sign must be reversed.

ax + b < c (or >, ≤, ≥)

Linear Equation in Two Variables

An equation that can be written in the form Ax + By = C, where A, B, and C are real numbers, and A and B are not both zero. Its graph is a straight line on a coordinate plane. The slope-intercept form, y = mx + b, is often used for graphing, where m is the slope and b is the y-intercept.

Ax + By = C or y = mx + b

Linear Inequality in Two Variables

An inequality that can be written in the form Ax + By < C, Ax + By > C, Ax + By ≤ C, or Ax + By ≥ C. Its graph is a region on a coordinate plane, bounded by a straight line. The boundary line is solid for ≤ or ≥ (inclusive) and dashed for < or > (exclusive). The solution region is determined by testing a point.

Ax + By < C (or >, ≤, ≥)

System of Linear Equations

A set of two or more linear equations with the same variables. The solution to a system of two linear equations is the ordered pair (x, y) that satisfies all equations in the system. Graphically, it is the point of intersection of the lines. Systems can have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (coincident lines).

y = m1x + b1 and y = m2x + b2 (example)

System of Linear Inequalities

A set of two or more linear inequalities with the same variables. The solution to a system of linear inequalities is the region on the coordinate plane where the shaded regions of all inequalities overlap.

y < m1x + b1 and y > m2x + b2 (example)

Key facts to remember

1To solve an equation, perform the same operation to both sides to isolate the variable.
2When multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality sign.
3The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.
4For linear inequalities in two variables, use a solid line for ≤ or ≥ and a dashed line for < or >.
5The solution to a system of linear equations is the point(s) where their graphs intersect.
6The solution to a system of linear inequalities is the region where the shaded areas of all inequalities overlap.
7Parallel lines have the same slope and never intersect (no solution for systems of equations).
8Perpendicular lines have slopes that are negative reciprocals of each other.

Worked examples

Example 1

Solve 3(x - 2) + 5 ≥ 2x + 7 and graph the solution on a number line.

I3x - 6 + 5 ≥ 2x + 7 (Distribute 3)

II3x - 1 ≥ 2x + 7 (Combine like terms)

III3x - 2x - 1 ≥ 7 (Subtract 2x from both sides)

IVx - 1 ≥ 7 (Simplify)

Vx ≥ 7 + 1 (Add 1 to both sides)

VIx ≥ 8 (Simplify)

VIIGraph: Draw a number line, place a closed circle at 8, and shade to the right.

Answer

x ≥ 8

Remember to distribute carefully and combine like terms before isolating the variable.

Example 2

Graph the equation y = -2x + 3 and the inequality y > (1/2)x - 1 on the same coordinate plane.

IFor y = -2x + 3:

II Identify y-intercept b = 3. Plot (0, 3).

III Identify slope m = -2 = -2/1. From (0, 3), go down 2 units and right 1 unit to (1, 1).

IV Draw a solid line through these points.

VFor y > (1/2)x - 1:

VI Identify y-intercept b = -1. Plot (0, -1).

VII Identify slope m = 1/2. From (0, -1), go up 1 unit and right 2 units to (2, 0).

VIII Draw a dashed line through these points because the inequality is >.

9 Choose a test point not on the line, e.g., (0, 0).

10 Substitute (0, 0) into y > (1/2)x - 1: 0 > (1/2)(0) - 1 which simplifies to 0 > -1. This is true.

11 Shade the region that contains (0, 0) (above the dashed line).

Answer

The graph shows a solid line for y = -2x + 3 passing through (0,3) and (1,1). It also shows a dashed line for y = (1/2)x - 1 passing through (0,-1) and (2,0), with the region above this dashed line shaded.

Pay close attention to whether the boundary line should be solid or dashed, and use a test point to determine the correct shading.

Example 3

Solve the system of equations using the substitution method:\ny = 2x + 1\n3x + 2y = 16

IThe first equation y = 2x + 1 is already solved for y.

IISubstitute (2x + 1) for y in the second equation: 3x + 2(2x + 1) = 16.

IIIDistribute: 3x + 4x + 2 = 16.

IVCombine like terms: 7x + 2 = 16.

VSubtract 2 from both sides: 7x = 14.

VIDivide by 7: x = 2.

VIISubstitute x = 2 back into the first equation to find y: y = 2(2) + 1.

VIIISimplify: y = 4 + 1, so y = 5.

9Check the solution (2, 5) in both original equations:

10 5 = 2(2) + 1 -> 5 = 4 + 1 -> 5 = 5 (True)

11 3(2) + 2(5) = 16 -> 6 + 10 = 16 -> 16 = 16 (True)

Answer

The solution is (2, 5).

Always check your solution in *both* original equations to ensure accuracy.

Common mistakes

✗Forgetting to reverse the inequality sign when multiplying or dividing by a negative number.
✗Making arithmetic errors, especially with negative numbers, during distribution or combining like terms.
✗Incorrectly drawing boundary lines for inequalities (using a solid line instead of dashed, or vice-versa).
✗Shading the wrong region for inequalities; always use a test point to verify.
✗Algebraic errors when substituting or eliminating variables in systems of equations.

Exam tips

★Always show all your steps clearly, even for simple calculations, to earn partial credit and help identify errors.
★When graphing, use graph paper and a ruler to ensure accuracy. Label your axes and scales.
★For systems of equations, check your solution by substituting the ordered pair back into *all* original equations.
★For systems of inequalities, clearly indicate the final solution region by shading it distinctly or using a different color.

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