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Q10.

Expert-verified

Algebra 2

Found in: Page 149

Book edition Middle English Edition

Author(s) Carter

Pages 804 pages

ISBN 9780079039903

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Short Answer

Solve each system of inequalities by graphing.
$\begin{matrix} y \geq x - 3 \\ y \geq - x + 1 \end{matrix}$

The solution of the inequalities is $- y + 1 \leq x \leq y + 3$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step-1 – Concept of solution of linear inequalities

The solution of linear inequalities can be obtained by changing the inequalities into equations and solving the linear equations to obtain a graph. Then the common shaded region is a solution of the inequalities.

Step-2 – Concept of shading the region of the inequalities

The shaded region obtained by choosing a point, if the point satisfies the inequality the region is along the point, if not satisfies the inequalities, then the shaded region is opposite to the point.

Step-3 – Solving the inequalities

The given inequalities are-:

$\begin{matrix} y \geq x - 3 \\ y \geq - x + 1 \end{matrix}$

The linear equation of the inequalities is-:

$\begin{matrix} y = x - 3 \\ y = - x + 1 \end{matrix}$

The point which satisfies the equation $y = x - 3$ are $(0, - 3)$ and $(3, 0)$ .

The point which satisfies the equation $y = - x + 1$ are $(0, 1)$ and $(1, 0)$ .

Step-4 – Evaluating the shading region

We choose $(0, 0)$ then the point $(0, 0)$ satisfies the inequality $y \geq x - 3$ but it does not satisfy the inequality $y \geq - x + 1$ .

Step-5 – Plotting the graph

So, the graph of the inequality is

The common region is $- y + 1 \leq x \leq y + 3$

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