Graphing a System of Two Linear Inequalities in Slope-Intercept Form
Step 1: Graph one of the inequalities as if it were an equation, {eq}y = mx + b{/eq}, using the {eq}y{/eq}-intercept {eq}(0,b){/eq} and the slope, {eq}m{/eq}. If the original inequality was {eq}\leq{/eq} or {eq}\geq{/eq}, use a solid line, and if the original inequality was {eq}<{/eq} or {eq}>{/eq}, use a dashed line.
Step 2: Pick a point on one side of the line graphed in step 1 and substitute the coordinates into the original inequality. If the result is a true statement, shade the side of the graph that contains that point. If the result is a false statement, shade the side of the graph that does not contain that point.
Step 3: Graph the other inequality on the same graph by repeating steps 1 and 2. The solution to the system, if it exists, consists of the region that is shaded by both graphs simultaneously.
Graphing a System of Two Linear Inequalities in Slope-Intercept Form - Vocabulary and Equations
Linear Inequality: A linear inequality is a mathematical sentence in which there are no exponents, and there is an inequality symbol ({eq}<, \ > , \ \leq , \ \geq{/eq}). Linear inequalities can be written in slope-intercept form, {eq}y < mx + b{/eq}, where the {eq}<{/eq} symbol can be any of the inequality symbols.
System of Linear Inequalities: A system of linear inequalities is made up of two or more linear inequalities. A solution to a system of linear inequalities is the set of all points which satisfy all of the inequalities simultaneously.
We will use these steps, definitions, and equations to identify the correct graph of a system of two linear inequalities in the slope-intercept form in the following two examples.
Example Problem 1: Graphing a System of Two Linear Inequalities in Slope-Intercept Form
Choose the correct graph of the system of linear inequalities below.
{eq}y \geq 3x - 1\\y < -x + 2{/eq}
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Step 1: Graph one of the inequalities as if it were an equation, {eq}y = mx + b{/eq}, using the {eq}y{/eq}-intercept {eq}(0,b){/eq} and the slope, {eq}m{/eq}. If the original inequality was {eq}\leq{/eq} or {eq}\geq{/eq}, use a solid line, and if the original inequality was {eq}<{/eq} or {eq}>{/eq}, use a dashed line.
To graph {eq}y \geq 3x - 1{/eq} as if it were an equation, {eq}y = 3x - 1{/eq}, we need to plot the {eq}y{/eq}-intercept {eq}(0,-1){/eq} and use a slope of {eq}3{/eq}, moving up 3 and right 1 from our {eq}y{/eq}-intercept and connecting the points. Since the original inequality was {eq}\geq{/eq}, we need to use a solid line.
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Step 2: Pick a point on one side of the line graphed in step 1 and substitute the coordinates into the original inequality. If the result is a true statement, shade the side of the graph that contains that point. If the result is a false statement, shade the side of the graph that does not contain that point.
We can pick any point we want to check in the inequality, but {eq}(0,0){/eq} is the easiest to substitute, so we will use it. Substituting into {eq}y\geq 3x - 1{/eq}, we have:
{eq}y\geq 3x - 1\\0\geq 3(0) - 1\\0\geq -1{/eq}
This is a true statement since 0 is greater than -1. So we need to shade the side of the graph that contains the point {eq}(0,0){/eq}.
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Step 3: Graph the other inequality on the same graph by repeating steps 1 and 2. The solution to the system, if it exists, consists of the region that is shaded by both graphs simultaneously.
To graph {eq}y < -x + 2{/eq}, first graph {eq}y = -x + 2{/eq} with a dashed line, using the {eq}y{/eq}-intercept {eq}(0,2){/eq} and slope, {eq}-1{/eq}.
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Checking the point {eq}(0,0){/eq} again, we have:
{eq}y < -x + 2\\0 < -0 + 2\\0 < 2{/eq}
This is a true statement, and so we shade the side of the graph that contains {eq}(0,0){/eq}.
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This matches graph B.
Example Problem 2: Graphing a System of Two Linear Inequalities in Slope-Intercept Form
Choose the correct graph of the system of linear inequalities below.
{eq}y < \dfrac{1}{2}x - 3\\y > 2x + 2{/eq}
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Step 1: Graph one of the inequalities as if it were an equation, {eq}y = mx + b{/eq}, using the {eq}y{/eq}-intercept {eq}(0,b){/eq} and the slope, {eq}m{/eq}. If the original inequality was {eq}\leq{/eq} or {eq}\geq{/eq}, use a solid line, and if the original inequality was {eq}<{/eq} or {eq}>{/eq}, use a dashed line.
Graph {eq}y <\dfrac{1}{2}x + 3{/eq} using a dashed line, the {eq}y{/eq}-intercept {eq}(0,3){/eq}, and slope, {eq}\dfrac{1}{2}{/eq}.
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Step 2: Pick a point on one side of the line graphed in step 1 and substitute the coordinates into the original inequality. If the result is a true statement, shade the side of the graph that contains that point. If the result is a false statement, shade the side of the graph that does not contain that point.
Substituting in {eq}(0,0){/eq}, we have:
{eq}y < \dfrac{1}{2}x + 3\\0 < \dfrac{1}{2}(0) + 3\\0 < 3{/eq}
This is a true statement, and so we shade that side of the graph.
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Step 3: Graph the other inequality on the same graph by repeating steps 1 and 2. The solution to the system, if it exists, consists of the region that is shaded by both graphs simultaneously.
To graph {eq}y > 2x + 2{/eq}, use a dashed line, {eq}y{/eq}-intercept {eq}(0,2){/eq}, and slope, {eq}2{/eq}.
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Checking the point {eq}(0,0){/eq}, we have:
{eq}y > 2x + 2\\0 > 2(0) + 2\\0 > 2{/eq}
This is a false statement, and so we shade the side of the graph that does not contain {eq}(0,0){/eq}.
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This matches graph C.
