Chapter 9
Algebra and Trigonometry with Analytic Geometry · 264 exercises
Problem 1
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{8 x-1}{(x-2)(x+3)} $$
7 step solution
Problem 1
Exer. 1-14: Without expanding, explain why the statement is true. $$ \left|\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array}\right|=-\left|\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right| $$
4 step solution
Problem 1
Exer. 1-2: Show that \(B\) is the inverse of \(A\). $$ A=\left[\begin{array}{ll} 5 & 7 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & -7 \\ -2 & 5 \end{array}\right] $$
5 step solution
Problem 1
Find, if possible, \(A+B, A-B, 2 A\), and \(-3 B\) $$ A=\left[\begin{array}{rr} 5 & -2 \\ 1 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 4 & 1 \\ -3 & 2 \end{array}\right] $$
5 step solution
Problem 1
\(\left\\{\begin{array}{rr}x-2 y-3 z= & -1 \\ 2 x+y+z= & 6 \\ x+3 y-2 z= & 13\end{array}\right.\)
6 step solution
Problem 1
Exer. 1-22: Solve the system. $$ \left\\{\begin{array}{r} 2 x+3 y=2 \\ x-2 y=8 \end{array}\right. $$
6 step solution
Problem 1
1-30: Use the method of substitution to solve the system. \(\left\\{\begin{array}{l}y=x^{2}-4 \\ y=2 x-1\end{array}\right.\)
6 step solution
Problem 2
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{x-29}{(x-4)(x+1)} $$
6 step solution
Problem 2
Exer. 1-2: Show that \(B\) is the inverse of \(A\). $$ A=\left[\begin{array}{rr} 8 & -5 \\ -3 & 2 \end{array}\right], \quad B=\left[\begin{array}{ll} 2 & 5 \\ 3 & 8 \end{array}\right] $$
4 step solution
Problem 2
Find, if possible, \(A+B, A-B, 2 A\), and \(-3 B\) $$ A=\left[\begin{array}{rr} 3 & 0 \\ -1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & -4 \\ 1 & 1 \end{array}\right] $$
4 step solution
Problem 2
\(\left\\{\begin{array}{rr}x+3 y-z= & -3 \\ 3 x-y+2 z= & 1 \\ 2 x-y+z= & -1\end{array}\right.\)
7 step solution
Problem 2
Solve the system. $$ \left\\{\begin{array}{l} 4 x+5 y=13 \\ 3 x+y=-4 \end{array}\right. $$
5 step solution
Problem 2
1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{aligned} y &=x^{2}+1 \\ x+y &=3 \end{aligned}\right. $$
5 step solution
Problem 3
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{x+34}{x^{2}-4 x-12} $$
6 step solution
Problem 3
Exer. 3-12: Find the inverse of the matrix if it exists. $$ \left[\begin{array}{rr} 2 & -4 \\ 1 & 3 \end{array}\right] $$
5 step solution
Problem 3
\(\left\\{\begin{array}{rr}5 x+2 y-z= & -7 \\ x-2 y+2 z= & 0 \\ 3 y+z= & 17\end{array}\right.\)
5 step solution
Problem 3
Find, if possible, \(A+B, A-B, 2 A\), and \(-3 B\) $$ A=\left[\begin{array}{rr} 6 & -1 \\ 2 & 0 \\ -3 & 4 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & 1 \\ -1 & 5 \\ 6 & 0 \end{array}\right] $$
5 step solution
Problem 3
Exer. 3-4: Sketch the region \(R\) determined by the given constraints, and label its vertices. Find the maximum value of \(C\) on \(R\). $$ \begin{array}{ll} C=3 x+y ; & x \geq 0, y \geq 0, \\ 3 x-4 y \geq-12, & 3 x+2 y \leq 24, \quad 3 x-y \leq 15 \end{array} $$
6 step solution
Problem 3
\(2 x+3 y \geq 2 y+1\)
3 step solution
Problem 3
Solve the system. $$ \left\\{\begin{array}{l} 2 x+5 y=16 \\ 3 x-7 y=24 \end{array}\right. $$
5 step solution
Problem 3
1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{aligned} y^{2} &=1-x \\ x+2 y &=1 \end{aligned}\right. $$
5 step solution
Problem 4
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{5 x-12}{x^{2}-4 x} $$
7 step solution
Problem 4
Exer. 1-14: Without expanding, explain why the statement is true. $$ \left|\begin{array}{lll} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{array}\right|=\left|\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{array}\right| $$
4 step solution
Problem 4
Find the inverse of the matrix if it exists. $$ \left[\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right] $$
5 step solution
Problem 4
Find, if possible, \(A+B, A-B, 2 A\), and \(-3 B\) $$ A=\left[\begin{array}{rrr} 0 & -2 & 7 \\ 5 & 4 & -3 \end{array}\right], \quad B=\left[\begin{array}{lll} 8 & 4 & 0 \\ 0 & 1 & 4 \end{array}\right] $$
5 step solution
Problem 4
\(\left\\{\begin{aligned} 4 x-y+3 z &=6 \\\\-8 x+3 y-5 z &=-6 \\ 5 x-4 y &=-9 \end{aligned}\right.\)
7 step solution
Problem 4
Exer. 3-4: Sketch the region \(R\) determined by the given constraints, and label its vertices. Find the maximum value of \(C\) on \(R\). $$ \begin{aligned} &C=4 x-2 y \\ &x-2 y \geq-8, \quad 7 x-2 y \leq 28, \quad x+y \geq 4 \end{aligned} $$
6 step solution
Problem 4
1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{r} y^{2}=x \\ x+2 y+3=0 \end{array}\right. $$
4 step solution
Problem 4
Solve the system. $$ \left\\{\begin{array}{l} 7 x-8 y=9 \\ 4 x+3 y=-10 \end{array}\right. $$
8 step solution
Problem 5
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{4 x^{2}-15 x-1}{(x-1)(x+2)(x-3)} $$
7 step solution
Problem 5
Exer. 1-14: Without expanding, explain why the statement is true. $$ \left|\begin{array}{lll} 2 & 4 & 2 \\ 1 & 2 & 4 \\ 2 & 6 & 4 \end{array}\right|=4\left|\begin{array}{lll} 1 & 2 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 2 \end{array}\right| $$
4 step solution
Problem 5
Find the inverse of the matrix if it exists. $$ \left[\begin{array}{ll} 2 & 4 \\ 4 & 8 \end{array}\right] $$
2 step solution
Problem 5
Find, if possible, \(A+B, A-B, 2 A\), and \(-3 B\) $$ A=\left[\begin{array}{lll} 4 & -3 & 2 \end{array}\right], \quad B=\left[\begin{array}{lll} 7 & 0 & -5 \end{array}\right] $$
5 step solution
Problem 5
\(\left\\{\begin{array}{rr}2 x+6 y-4 z= & 1 \\ x+3 y-2 z= & 4 \\ 2 x+y-3 z= & -7\end{array}\right.\)
4 step solution
Problem 5
Exer. 5-6: Sketch the region \(R\) determined by the given constraints, and label its vertices. Find the minimum value of \(C\) on \(R\). $$ \begin{array}{ll} C=3 x+6 y ; & x \geq 0, y \geq 0 \\ 2 x+3 y \geq 12, & 2 x+5 y \geq 16 \end{array} $$
6 step solution
Problem 5
1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{aligned} 2 y &=x^{2} \\ y &=4 x^{3} \end{aligned}\right. $$
6 step solution
Problem 5
Solve the system. $$ \left\\{\begin{aligned} 3 r+4 s &=3 \\ r-2 s &=-4 \end{aligned}\right. $$
6 step solution
Problem 6
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{x^{2}+19 x+20}{x(x+2)(x-5)} $$
6 step solution
Problem 6
Find the inverse of the matrix if it exists. $$ \left[\begin{array}{ll} 3 & -1 \\ 6 & -2 \end{array}\right] $$
3 step solution
Problem 6
Find, if possible, \(A+B, A-B, 2 A\), and \(-3 B\) $$ A=\left[\begin{array}{r} 7 \\ -16 \end{array}\right], \quad B=\left[\begin{array}{r} -11 \\ 9 \end{array}\right] $$
4 step solution
Problem 6
\(\left\\{\begin{array}{r}x+3 y-3 z=-5 \\ 2 x-y+z=-3 \\ -6 x+3 y-3 z=4\end{array}\right.\)
6 step solution
Problem 6
Exer. 5-6: Sketch the region \(R\) determined by the given constraints, and label its vertices. Find the minimum value of \(C\) on \(R\). $$ \begin{array}{ll} C=6 x+y ; & y \geq 0 \\ 3 x+y \geq 3, & x+5 y \leq 15, \quad 2 x+y \leq 12 \end{array} $$
6 step solution
Problem 6
\( y^{2}-x \leq 0\)
4 step solution
Problem 6
1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{aligned} x-y^{3} &=1 \\ 2 x &=9 y^{2}+2 \end{aligned}\right. $$
7 step solution
Problem 6
Solve the system. $$ \left\\{\begin{array}{l} 9 u+2 v=0 \\ 3 u-5 v=17 \end{array}\right. $$
5 step solution
Problem 7
Exer. 1-28: Find the partial fraction decomposition. $$ \frac{4 x^{2}-5 x-15}{x^{3}-4 x^{2}-5 x} $$
6 step solution
Problem 7
Exer. 1-14: Without expanding, explain why the statement is true. $$ \left|\begin{array}{rrr} 1 & -1 & 2 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right|=0 $$
4 step solution
Problem 7
Find the inverse of the matrix if it exists. $$ \left[\begin{array}{rrr} 3 & -1 & 0 \\ 2 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right] $$
6 step solution
Problem 7
\(\left\\{\begin{array}{rr}2 x-3 y+2 z= & -3 \\ -3 x+2 y+z= & 1 \\ 4 x+y-3 z= & 4\end{array}\right.\)
6 step solution
Problem 7
Exer. 7-8: Sketch the region \(R\) determined by the given constraints, and label its vertices. Describe the set of points for which \(C\) is a maximum on \(R\). $$ \begin{array}{lll} C=2 x+4 y ; & x \geq 0, y \geq 0, & \\ x-2 y \geq-8, & \frac{1}{2} x+y \leq 6, & 3 x+2 y \leq 24 \end{array} $$
4 step solution