/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Free solutions & answers for Elementary and Intermediate Algebra Chapter 4 - (Page 32) [step by step] | 91Ó°ÊÓ

91Ó°ÊÓ

Problem 87

Solve each inequality. Write the solution set in interval notation and graph it. $$ -4(2 y+2) \leq 4 y+28 $$

Problem 87

What concept does this diagram illustrate? $$ \left\\{\begin{array}{l} {6 x+5 y=11} \\ {y=(3 x-2)} \end{array}\right. $$

Problem 88

Solve each inequality. Write the solution set in interval notation and graph it. $$ -5<3 t+4 \leq 13 $$

Problem 88

Explain the error. Solve: \(\quad\left\\{\begin{array}{l}{3 a+4 b=1} \\ {a+2 b=9}\end{array}\right.\) \(a=9-2 b \quad\) Solve for \(a\) in the second equation. \(9-2 b+2 b=9 \quad\) Substitute \(9-2 b\) for \(a\) \(\begin{aligned} 9=9 & \text { The b-terms drop out. } \\ \text { True } \end{aligned}\) The system has infinitely many solutions.

Problem 88

Solve the system by either the substitution or the elimination method. $$ \left\\{\begin{array}{l} {11 c+3 d=-68} \\ {10 c+3 d=-64} \end{array}\right. $$

Problem 89

When using the substitution method, how can you tell whether a. a system of linear equations has no solution? b. a system of linear equations has infinitely many solutions?

Problem 89

Solve the system by either the substitution or the elimination method. $$ \left\\{\begin{array}{l} {\frac{2}{15} x-\frac{1}{5} y=\frac{1}{3}} \\ {\frac{2}{15} x-\frac{1}{5} y=\frac{1}{10}} \end{array}\right. $$

Problem 89

Solve each inequality. Write the solution set in interval notation and graph it. $$ -1 \leq-\frac{1}{2} n $$

Problem 90

Solve each inequality. Write the solution set in interval notation and graph it. $$ \frac{1}{3}+\frac{c}{5}>-\frac{3}{2} $$

Problem 90

Solve the system by either the substitution or the elimination method. $$ \left\\{\begin{array}{l} {\frac{1}{5} x+\frac{3}{5} y=\frac{4}{5}} \\ {\frac{1}{6} x+\frac{1}{2} y=\frac{2}{3}} \end{array}\right. $$

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