In Exercises
lineareqfirst -
lineareqlast , solve the given linear equation and check your answer.
\(\frac {3 - 2t}{4} = 7t+1\)
\(t = \answer {-\frac {1}{30}}\)
\(\frac {2(w-3)}{5} = \frac {4}{15} - \frac {3w+1}{9}\)
\(-0.02y + 1000 = 0\)
\(y = \answer {50000}\)
\(\frac {49w - 14}{7}= 3w - (2-4w)\)
\(7 - (4-x) = \frac {2x-3}{2}\)
\(3 t\sqrt {7} + 5 = 0\)
\(t = -\frac {5}{3\sqrt {7}} = -\frac {5\sqrt {7}}{21}\)
\(\sqrt {50} y = \frac {6 - \sqrt {8} y}{3}\)
\(y = \frac {6}{17\sqrt {2}} = \frac {3 \sqrt {2}}{17}\)
\(4 - (2x+1) = \frac {x \sqrt {7}}{9}\)
\(x = \frac {27}{18+\sqrt {7}}\)
In equations literalexfirst - literalexlast , solve each equation for the indicated variable.
Solve for
\(y\) :
\(3x+2y = 4\)
\(y = \frac {4 - 3x}{2}\) or \(y = -\frac {3}{2}x + 2\)
Solve for
\(x\) :
\(3x+2y = 4\)
\(x = \frac {4 - 2y}{3}\) or \(x = -\frac {2}{3} y + \frac {4}{3}\)
Solve for
\(C\) :
\(F = \frac {9}{5} C + 32\)
\(C = \frac {5}{9}(F - 32)\) or \(C = \frac {5}{9} F - \frac {160}{9}\)
Solve for
\(x\) :
\(p = -2.5x + 15\)
\(x = \frac {p - 15}{-2.5} = \frac {15-p}{2.5}\) or \(x = -\frac {2}{5} p + 6\) .
Solve for
\(x\) :
\(C = 200x + 1000\)
\(x = \frac {C - 1000}{200}\) or \(x = \frac {1}{200} C - 5\)
Solve for
\(y\) :
\(x= 4(y+1) + 3\)
\(y = \frac {x-7}{4}\) or \(y = \frac {1}{4} x - \frac {7}{4}\)
Solve for
\(w\) :
\(vw - 1 = 3v\)
\(w = \frac {3v+1}{v}\) , provided \(v \neq 0\)
Solve for
\(v\) :
\(vw - 1 = 3v\)
\(v = \frac {1}{w-3}\) , provided \(w \neq 3\)
Solve for
\(y\) :
\(x(y-3) = 2y+1\)
\(y = \frac {3x+1}{x-2}\) , provided \(x \neq 2\) .
Solve for
\(\pi \) :
\(C = 2\pi r\)
\(\pi = \frac {C}{2r}\) , provided \(r \neq 0\) .
Solve for
\(V\) :
\(PV = nRT\)
\(V = \frac {nRT}{P}\) , provided \(P \neq 0\) .
Solve for
\(R\) :
\(PV = nRT\)
\(R = \frac {PV}{nT}\) , provided \(n \neq 0, T \neq 0\) .
Solve for
\(g\) :
\(E = mgh\)
\(g = \frac {E}{mh}\) , provided \(m \neq 0, h \neq 0\) .
Solve for
\(m\) :
\(E = \frac {1}{2} mv^2\)
\(m = \frac {2E}{v^2}\) , provided \(v^2 \neq 0\) (so \(v \neq 0\) )
In Exercises subex1 - subex2 , the subscripts on the variables have no intrinsic mathematical meaning; they’re just used to distinguish one
variable from another. In other words, treat ‘\(P_{1}\) ’ and ‘\(P_{2}\) ’ as two different variables as you would ‘\(x\) ’ and ‘\(y\) .’ (The same goes for ‘\(x\) ’ and ‘\(x_{0}\) ,’
etc.)
Solve for
\(V_{2}\) :
\(P_{1}V_{1} = P_{2}V_{2}\)
\(V_{2} = \frac {P_{1}V_{1}}{P_{2}}\) , provided \(P_{2} \neq 0\) .
Solve for
\(t\) :
\(x = x_{0} + at\)
\(t = \frac {x - x_{0}}{a}\) , provided \(a \neq 0\) .
Solve for
\(x\) :
\(y-y_{0} = m(x -x_{0})\)
\(x = \frac {y-y_{0} + mx_{0}}{m}\) or \(x = x_{0} + \frac {y-y_{0}}{m}\) , provided \(m \neq 0\) .
Solve for
\(T_{1}\) :
\(q = mc(T_{2} -T_{1})\)
\(T_{1} = \frac {mcT_{2} - q}{mc}\) or \(T_{1} = T_{2} - \frac {q}{mc} \) , provided \(m \neq 0, c \neq 0\) .
With the help of your classmates, find values for
\(c\) so that the equation:
\(2x - 5c = 1 - c(x+2)\)
has \(x = 42\) as a solution.
has no solution (that is, the equation is a contradiction.)
Is it possible to find a value of \(c\) so the equation is an identity? Explain.
In Exercises linineqnexfirst - linineqnexlast , solve the given inequality. Write your answer using interval notation.
\(3 - 4x \geq 0\)
\(\left (-\infty , \frac {3}{4}\right ]\)
\(2t - 1 < 3 - (4t-3)\)
\(\left (\answer {-\infty }, \answer {\frac {7}{6}} \right )\)
\(\frac {7 -y}{4} \geq 3y + 1\)
\(\left ( -\infty , \frac {3}{13}\right ]\)
\(0.05R + 1.2 > 0.8 - 0.25R\)
\(\left (\answer {-\frac {4}{3}}, \answer {\infty }\right )\)
\(\frac {10m+1}{5} \geq 2m - \frac {1}{2}\)
\(\left (\answer {-\infty }, \answer {\infty } \right )\)
\(x \sqrt {12} - \sqrt {3} > \sqrt {3} x + \sqrt {27}\)
\(\left (\answer {4}, \answer {\infty }\right )\)
\(2t - 7 \leq \sqrt [3]{18} t\)
\(\left [ \frac {7}{2 - \sqrt [3]{18}}, \infty \right )\)
\(117y \geq y\sqrt {2} - 7y \sqrt [4]{8}\)
\(\left [\answer {0}, \answer {\infty }\right )\)
\(-\frac {1}{2} \leq 5x - 3 \leq \frac {1}{2}\)
\(\left [ \frac {1}{2}, \frac {7}{10}\right ]\)
\(-\frac {3}{2} \leq \frac {4 - 2t}{10} < \frac {7}{6}\)
\(\left (\answer {-\frac {23}{6}}, \answer {\frac {19}{2}} \right ]\)
\(-0.1 \leq \frac {5-x}{3} - 2 < 0.1\)
\(\left (-\frac {13}{10}, -\frac {7}{10} \right ]\)
\(2y \leq 3-y < 7\)
\(\left (\answer {-4},\answer {1} \right ]\)
\(6-5t > \frac {4t}{3} \geq t - 2\)
\(\left [\answer {-6}, \answer {\frac {18}{19}} \right )\)
\(2x+1 \leq -1\) or
\(2x+1 \geq 1\)
\((-\infty , -1] \cup [0, \infty )\)
\(4-x \leq 0\) or
\(2x+7 < x\)
\(\left (\answer {-\infty }, \answer {-7} \right ) \cup \left [ \answer {4}, \answer {\infty } \right )\)
\(\frac {5-2x}{3} > x\) or
\(2x + 5 \geq 1\)
