For real numbers \(a\) and \(b\) where \(a\neq 0\) the equation $$ a\cdot x = b$$ is called the linear equation or the equation of the first degree were the variable x is the only unknown. Solving the equations means to determine every real number \(x_1\)
for which the equality \(a\cdot x_1 = b\) is true. Then the \(x_1\) is named the solution of the equation.
- Example 1 - Solve the linear equation \(2x-2\{x-2\left[x-2(x-2)\right]\} = 0.\)
Solution$$\begin{eqnarray} 2x-2\{x-2\left[x-2(x-2)\right]\} &=& 0 \\ \nonumber 2x-2\{x-2\left[4-x\right]\} &=& 0 \\ \nonumber 2x-2\{3x-8\} &=& 0 \\ \nonumber 2x-6x+16 &=& 0 \\ \nonumber -4x+16 &=& 0 \\ \nonumber -4x &=& -16 \\ \nonumber x &=& 4\end{eqnarray}$$ - Example 2 - Solve the linear equation \(\frac{x-9}{4} + \frac{2x-1}{6} - \frac{3x+11}{9} = 3\)
Solution$$\begin{eqnarray}\frac{x-9}{4} + \frac{2x-1}{6} - \frac{3x+11}{9} &=& 3/\cdot 36 \\ \nonumber 9(x-9)+6(2x-1)-4(3x+11) &=& 108 \\\nonumber 9x-81 + 12x-6 - 12x-44 &=& 108 \\\nonumber 9x &=& 239 \\ \nonumber x &=& \frac{239}{9} \end{eqnarray}$$ - Example 3 - Solve the equation \(\frac{4}{x^2 + 10x +25} + \frac{3}{10x-2x^2} = \frac{5}{2x^2-50}\)
Solution $$\begin{eqnarray}\frac{4}{x^2 + 10x +25} + \frac{3}{10x-2x^2} &=& \frac{5}{2x^2-50}\\\nonumber \frac{4}{(x+5)^2} + \frac{3}{2x(5-x)} &=& \frac{5}{2(x-5)(x+5)}/\cdot 2x(x-5)(x+5)^2 \\ \nonumber 4\cdot 2x(x-5) -3(x+5)^2 &=& 5x(x+5) \\\nonumber 8x^2 - 40x - 3(x^2 +10x+25) &=& 5x^2 +25 \\ \nonumber 8^2 -40x - 3x^2 -30x-75 &=& 5x^2 + 25x \\ \nonumber -95x &=& 75 \\\nonumber x &=& -\frac{15}{19}\end{eqnarray}$$ - Example 4 - Solve the equation \(\frac{1}{2}m(x+1) - \frac{1}{2}(2x-1) = 1\) with a discussion of the dependence of the solution on the real parameter m.
Solution $$\begin{eqnarray} \frac{1}{2}m(x+1) - \frac{1}{2}(2x-1) &=& 1 /\cdot 2 \\ \nonumber mx + m -2x+1 &=& 2 \\ \nonumber x(m-2) &=& 1-n \end{eqnarray}$$ The number that multiplies \(x\) can be equal zero or different from zero so we need to investigate these two cases. First case - the \(m-2 = 0\) $$ m -2 = 0 \Rightarrow m = 2 \Rightarrow x\cdot 0 = -1.$$ The previous expression is not valid. In second case the \(m - 2 \neq 0\) i.e. $$ m - 2 \neq 0 \Rightarrow m\neq 2 $$ $$x = \frac{1-m}{m-2} $$ The solution of the equation is \(x = \frac{1-m}{m-2}\) for every real number except for the number 2. - Example 5 - Solve the equation \(\frac{x}{x-1} + \frac{x-2}{x} = 2-\frac{2}{x}\)
Solution $$\begin{eqnarray} \frac{x}{x-1} + \frac{x-2}{x} &=& 2-\frac{2}{x} /\cdot x(x-1)\end{eqnarray}$$ The multiplication of the linear equation can be done with condition that \(x\neq 1\) and \(x\neq 0\) $$\begin{eqnarray}x^2+(x-2)(x-1) &=& 2x(x-1)-2(x-1) \\\nonumber x^2 + x^2 -x - 2x + 2 &=& 2x^2 -2x - 2x + 2 \\ \nonumber -3x &=& -4x \\ \nonumber x &=& 0\end{eqnarray}$$ The obtained solution is not the solution due to the condition \(x\neq 0.\) - Example 6 - If \(O = 2r\pi(r+v),\) determine how much is the \(v?\)
Solution $$ O = 2r\pi(r+v)/:2r\pi \Rightarrow \frac{O}{2r\pi} = r+v \Rightarrow v = \frac{O}{2r\pi}-r $$ - Example 7 - If \(P=\frac{a+c}{2}\cdot v,\) determine the value of variable \(v.\)
Solution $$\begin{eqnarray} P &=& \frac{a+c}{2}\cdot v/\cdot \frac{2}{a+c} \\ \nonumber \frac{2P}{a+c} &=& v \\\nonumber v&=& \frac{2P}{a+c}\end{eqnarray}$$ - Example 8 - Solve the equation \((x-4)(3+x) = 1 + (x-3)^2\)
Solution $$\begin{eqnarray}(x-4)(3+x) &=& 1 + (x-3)^2 \\ \nonumber 3x+x^2-12 - 4x &=& 1+ x^2 - 6x + 9 \\ \nonumber x^2 +3x-12 &=& x^2-6x+10 \\ \nonumber 5x &=&22 \\ \nonumber x = \frac{22}{5}\end{eqnarray}$$ - Example 9 - The equation \(2k + 5x + 3 = 0\) has the negative solution for which \(k ? \)
Solution First let's express the variable \(x\) $$ \begin{eqnarray}2k+5x+3 &=& 0 \\ \nonumber 5x &=& -2k-3/\cdot \frac{1}{5} \\ \nonumber x &=& -\frac{2k+3}{5}\end{eqnarray}$$ $$2k+3 \neq 0 \Rightarrow k\neq - \frac{3}{2} $$ If \(k < -\frac{3}{2}\) let 's say \(k = -\frac{5}{2}\) then the \(x\) is positive $$ x = \frac{-2\cdot \left(-\frac{5}{2}\right) - 3}{5} = \frac{2}{5}.$$ If \(k>-\frac{3}{2}\) let's say \(k=1 0\) then the \(x\) is negative $$ x=\ frac{-2\cdot 10 - 3}{5}=- \frac{23}{5}$$ - Example 10 - If \(O = 4a\) then \(a\) is equal to:
Solution $$ a = \frac{O}{4}.$$ - Example 11 - The solution to the equation \(1+x = 2(x-1)\)
Solution $$ \begin{eqnarray}1+x &=& 2(x-1) \\ \nonumber 1+x &=& 2x-2 \\ \nonumber -x &=& -3/\cdot (-1) \\ \nonumber x &=& 3\end{eqnarray}$$ - Example 12 - If \(p = \frac{ar+bs-q}{t}\) then \(s\) is equal to:
Solution $$ \begin{eqnarray}p &=& \frac{ar+bs-q}{t} /\cdot t \\ \nonumber pt &=& ar + bs - q \\ \nonumber bs &=& pt-ar+q/\cdot b \\ \nonumber s &=& \frac{pt-ar+q}{b}\end{eqnarray}$$ - Example 13 - If \(P = 6\) and if \(P = \frac{a+c}{2}\cdot v\) then how much is \(a+c\)
Solution $$\begin{eqnarray}P &=& \frac{a+c}{2}\cdot v/\cdot \frac{2}{v} \\ \nonumber \frac{2P}{v} &=& a+c \\ \nonumber a+c &=&\frac{12}{v}\end{eqnarray} $$ - Example 14 - Solve the equation \(-5+4(x-2) = 19-4x\)
Solution $$\begin{eqnarray}-5+4(x-2) &=& 19-4x \\ \nonumber -5+4x-8 &=& 19-4x \\ \nonumber 8x &=& 32 /\cdot \frac{1}{8} \\ \nonumber x &=& 4\end{eqnarray}$$ - Example 15 - Solve the equation \((x-1)(x+5) = x^2\)
Solution$$\begin{eqnarray}(x-1)(x+5) &=& x^2 \\ \nonumber x^2 +5x -x-5 &=& x^2 \\ \nonumber 4x &=& 5/\cdot \\ \nonumber x &=& \frac{5}{4}\end{eqnarray}$$ - Example 16 - Express the variable \(a\) from the expression \(p = ab+ (a+b)\cdot v\)
Solution $$\begin{eqnarray}p &=& ab+ (a+b)\cdot v \\ \nonumber p-bv &=& a(b+v)/\cdot \frac{1}{b+v} \\ \nonumber a &=& \frac{p-bv}{b+v}\end{eqnarray}$$ - Example 17 - Solve the equation \(5x - \frac{1}{2} = \frac{5}{2}-x\)
Solution $$\begin{eqnarray}5x - \frac{1}{2} &=& \frac{5}{2}-x/\cdot 2 \\\nonumber 10x -1 &=& 5-2x \\ \nonumber 8x &=& 6/\cdot 8 \\ \nonumber x &=& \frac{6}{8} \\ \nonumber x &=& \frac{3}{4}\end{eqnarray} $$ - Example 18 - Determine the \(h\) from the expression \(S = r\pi(r+2h)\)
Solution $$\begin{eqnarray} S &=& r\pi(r+2h) \\\nonumber S &=& r^2\pi + 2r\pi h \\ \nonumber 2r\pi h &=& S-r^2\pi/\cdot 2r\pi\\\nonumber h &=& \frac{S-r^2\pi}{2r\pi}\end{eqnarray} $$ - Example 19 - Solve the equation \(\frac{2x-3}{x+5} = -2\)
Solution $$\begin{eqnarray}\frac{2x-3}{x+5} &=& -2/\cdot (x+5)\\ \nonumber 2x-3 &=& -2x - 10 \\\nonumber 4x &=& -7 /\cdot 4 \\ \nonumber x &=& -\frac{7}{4}\end{eqnarray}$$ - Example 20 - If\(9x + 3y - 4 =0\) determine the y ?
Solution $$\begin{eqnarray} 9x+3y-4 &=& 0 \\ \nonumber 3y &=& -9x +4 /\cdot \frac{1}{3} \\ \nonumber y &=& -3x+\frac{4}{3}\end{eqnarray}$$ - Example 21 - Solve the equation \(\frac{2x+1}{x+1} + \frac{2x-1}{1-x} = \frac{-3}{x^2-1}\)
Solution $$ \begin{eqnarray}\frac{2x+1}{x+1} + \frac{2x-1}{1-x} &=& \frac{-3}{x^2-1}/\cdot (x-1)(x+1) \\ \nonumber (x-1)(2x+1) - (2x-1)(x+1) &=& -3 \\ \nonumber 2x^2 +x -2x -1 -2x^2 -2x +x + 1 &=&-3 \\ \nonumber -2x &=& -3/:(-2) \\ \nonumber x &=& \frac{3}{2}\end{eqnarray}$$ - Example 21 - Solve the equation \(\frac{4}{x^2+10x+25} + \frac{3}{10x-2x^2} = \frac{5}{2x^2-50}\)
Solution $$\begin{eqnarray}\frac{4}{x^2+10x+25} + \frac{3}{10x-2x^2} &=& \frac{5}{2x^2-50}/\cdot 2x(x-5)(x+5)^2\\\nonumber 4\cdot 2x(x-5) -3 \cdot (x+5)^2 &=& 5x(x+5)\\ \nonumber 8x^2-40x -3x^2 -30x - 75 &=& 5x^2 +25x \\\nonumber -95x &=& 75 \\ \nonumber x &=& -\frac{75}{95} \\ \nonumber x &=& -\frac{15}{19}\end{eqnarray}$$ - Example 22 - Solve the equation \(\frac{3}{x^2-9} - \frac{1}{9-6x+x^2} = \frac{4}{2x^2+6x}\)
Solution $$\begin{eqnarray}\frac{3}{x^2-9} - \frac{1}{9-6x+x^2} &=& \frac{4}{2x^2+6x}/\cdot 2x(x-3)^2(x+3) \\\nonumber 3\cdot 2x(x-3) - 2x(x+3) &=& 4(x-3)^2 \\ \nonumber 6x^2 - 18x - 2x^2 -6x &=& 4x^2 - 24x + 36\\\nonumber 0x &\neq& 36 \end{eqnarray}$$ The equation has no solution. - Example 23 - Determine the value of parameter \(m\) so that equation \(\frac{1}{2}m(x+1) - \frac{1}{2}(2x-1) = 1\) has no solution.
Solution $$\begin{eqnarray}\frac{1}{2}m(x+1) - \frac{1}{2}(2x-1) &=& 1 \\ \nonumber mx+m - 2x+1 &=& 2 \\ \nonumber mx+2x &=& 1-m \\ \nonumber x(m+2) &=& 1-m \\ \nonumber x &=& \frac{1-m}{m+2} \end{eqnarray}$$ If \(m = -1\) then \(x = \frac{1+1}{-1+2} = 2.\) If \(m = \frac{1}{2}\) then \(x=\frac{1-\frac{1}{2}}{\frac{1}{2} + 2} = \frac{\frac{1}{2}}{\frac{5}{2}} = \frac{1}{5}. \) If \(m = 2\) then \(x = \frac{1-2}{2+2} = -\frac{1}{2}\) $$\frac{1}{2}\cdot 2 \cdot (x+1) - \frac{1}{2}(2x-1) = 1 \Rightarrow x+1 - x-\frac{1}{2} = 1 \Rightarrow 0x = \frac{1}{2}$$If \(m=0\) then \(x = \frac{1-0}{0+2} = \frac{1}{2}\) - Example 24 - If \(m=0\) determine the solution of the equation \(2m(x-2)+2 = (2m-1)^2\)
Solution $$ \begin{eqnarray}2m(x-2)+2 &=& (2m-1)^2 \\ \nonumber 2m(x-2)+2 &=& 4m^2-4m+1\\ \nonumber x-2 &=& \frac{4m^2-4m-1}{2m} \\ \nonumber x &=& \frac{4m^2-1}{2m} \\\nonumber x &=& \frac{-1}{0} \Rightarrow \mathrm{No}\quad \mathrm{solution}\end{eqnarray}$$ - Example 25 - Determine how much solutions does the equation \(\frac{x-2}{x+2} - \frac{x+2}{x-2} = \frac{16}{x^2-4}\)
Solution $$\begin{eqnarray}\frac{x-2}{x+2} - \frac{x+2}{x-2} &=& \frac{16}{x^2-4}/\cdot (x-2)(x+2) \\ \nonumber (x-2)^2 - (x+2)^2 &=& 16 \\ \nonumber x^2-4x+4 - x^2-4x-4 &=& 16 \\\nonumber -8x &=& 16 \\\nonumber x &=& -2\end{eqnarray}$$ The equation has one solution and the solution is \(x = -2.\) - Example 26 - If \(O = r\pi(r+s)\) then \(s\) is equal to?
Solution $$ \begin{eqnarray}O &=& r\pi(r+s)/:r\pi \\ \nonumber \frac{O}{r\pi} -r &=& s \\ \nonumber s &=& \frac{O}{r\pi}-r\end{eqnarray}$$ - Example 27 - If \(y = \frac{x-1}{x+1}\) determine the value of variable \(x.\)
Solution $$\begin{eqnarray}y &=& \frac{x-1}{x+1}/\cdot (x+1) \\\nonumber y(x+1) &=& x-1\\ \nonumber yx+y - x + 1 &=& 0 \\ \nonumber x(y-1) &=& -(y+1)/\cdot\frac{1}{y-1}\\\nonumber x &=& -\frac{y+1}{y-1} \\ \nonumber x &=& \frac{y+1}{1-y}\end{eqnarray}$$ - Example 28 - Solve the equation \(8-2(3x+4) = 5x-16\)
Solution $$\begin{eqnarray}8-2(3x+4) &=& 5x-16 \\ \nonumber -6x-8-5x &=& -24 \\\nonumber -11x &=& -16/: (-11)\\ \nonumber x &=& \frac{16}{11}\end{eqnarray}$$ - Example 29 - The solution of the equation \(\frac{2}{x} + \frac{x}{x-2} = 1\) is?
Solution $$\begin{eqnarray}\frac{2}{x} + \frac{x}{x-2} &=& 1/\cdot x(x-2) \\ \nonumber 2(x-2) +x^2 &=& x(x-2) \\ \nonumber 2x-4 + x^2 &=& x^2 - 2x \\ \nonumber 4x &=& 4 \\ \nonumber x &=& 1\end{eqnarray}$$ Answer: The solution of the equation is \(x=1.\)
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