How to solve system of linear equations ?

• Example 1 - Solve the system of equations \(\begin{cases}2x& -5y = -5\\ 4x& + 3y = 3\end{cases}\).
  Solution The first equation is multiplied by -2 and then both equations are summed up. $$\begin{eqnarray}\begin{cases}2x -5y &=& -5/\cdot (-2)\\ 4x + 3y &=& 3\end{cases}\\ \nonumber \begin{cases}-4x +10y &=& 10\\ 4x + 3y &=& 3\end{cases}\Bigg\}+ \\ \nonumber -4x + 4x + 10y + 3y &=& 10 + 3\\ \nonumber 13y &=& 13/:13 \Rightarrow y = 1\end{eqnarray}$$ To find the value of variable \(x\) insert the \(y\) into the first or second equation. Here the \(y\) is inserted into the first equation. $$2x-5y = -5 \Rightarrow 2x-5 = -5 \Rightarrow x = 0$$ So, the solution of the system is \(\left(x,y\right) = \left(0,1\right)\)
• Example 2 - Solve the system of equations \(\begin{cases}2x& - y = -2\\ 4x& +3y = \frac{33}{5}\end{cases}\)
  SolutionThe procedure is similar as in the previous example. Multiply the first equation by \(-2\) and then add both equations together. $$\begin{eqnarray}\begin{cases}2x - y &=& -2/\cdot (-2)\\ 4x +3y &=& \frac{33}{5}\end{cases} \\ \nonumber \begin{cases}-4x + 2y &=& 4\\ 4x +3y &=& \frac{33}{5}\end{cases}\Bigg\}+ \\ \nonumber 5y = 4 + \frac{33}{5}\\ \nonumber 5y = \frac{33+20}{5}/\cdot \frac{1}{5}\Rightarrow y = \frac{53}{25}\end{eqnarray}$$ Substitute the \(y\) into the first equation. $$ 2x - y = -2 \Rightarrow 2x - \frac{53}{25} = -2 \Rightarrow 2x = \frac{53 - 50}{25}/cdot \frac{1}{2} \Rightarrow x = \frac{3}{50}$$ Solution of the system is \(\left(\frac{3}{50}, \frac{53}{25}\right)\)
• Example 3 - Solve the system of equations \(\begin{cases}4x& - 3y = 3 \\-8x& +6y = -6\end{cases}\)
  Solution - The first equation is multiplied by 2 and the equations are summed up. $$\begin{eqnarray}\begin{cases}4x-3y &=& 3/\cdot 2 \\ -8x+6y &=& -6\end{cases}\\ \nonumber \begin{cases}8x-6y &=& 6 \\ -8x + 6y &=& -6\end{cases}\Bigg\}+ 0&=&0 \end{eqnarray}$$ The system has infinite number of solutions (undetermined system).
• Example 4 -
  Solution
• Example 5 - Solve the system of eqations \(\begin{cases}4x + 5y &= 20 \\y = \frac{1}{2}x-2\end{cases}\)
  Solution - In this case the y variable is expressed over the x variable so the y variable will be substituted in the first equation. $$\begin{eqnarray} 4x + 5\left(\frac{1}{2}x-2\right) &=& 20 \\ \nonumber 4x + \frac{5}{2}x - 10 &=& 20 /\cdot 2 \\ \nonumber 8x + 5x - 20 &=& 40 \\ \nonumber 13x &=& 60\\\nonumber x &=& \frac{60}{13}\end{eqnarray}$$ Now the value of x is inserted into the second equation to determine the value of \(y\) variable. $$y = \frac{1}{2}x - 2 \Rightarrow y = \frac{1}{2}\frac{60}{13} - 2 \Rightarrow y = \frac{30-26}{13} \Rightarrow y = \frac{4}{13}$$
• Example 6 - Solve the system of equations \(\begin{cases}2x+5y &= -1 \\ 3x - 4y &= 33\end{cases}\)
  Solution - The first equation is multiplied with (-3) while the second equation is multiplied with (2). Then both eqatuions are summed up. $$\begin{eqnarray}\begin{cases}2x+5y &=& -1/\cdot (-3) \\ 3x - 4y &=& 33/\cdot 2\end{cases}\\ \nonumber \begin{cases}-6x-15y &=& 3 \\ 6x-8y &=& 66 \end{cases}\Bigg\}+ \\ \nonumber -15y - 8y &=& 3+66 \\ \nonumber -23y &=& 69 /:(-23) \\ \nonumber y &=& -3\end{eqnarray}$$ To determine the value of \(x\) insert the value of \(y\) into the first equation (or second). $$\begin{eqnarray}2x + 5y &=& -1 \\ \nonumber 2x+5\cdot(-3) &=& -1 \\ \nonumber 2x -15 &=& -1 \\\nonumber 2x &=& 14 \Rightarrow x = 7 \end{eqnarray}$$
• Example 7 - Solve the system of equations \(\begin{cases}x-y & = 3 \\ 3x- y &= 4\end{cases}\)
  Solution - From the first equation express the variable x over the variable y. $$x = 3+ y $$ Insert the new equation into the second equation. $$\begin{eqnarray}3x-y &=& 4 \\ \nonumber 3(3+y) - y &=& 4 \\ \nonumber 9+3y - y &=& 4 \\ \nonumber 2y &=& -5 \\\nonumber y &=& -\frac{5}{2} \end{eqnarray}$$ Insert the value \(y\) into the equation \(x = 3 + y\) to obtaine the value of \(x\). $$\begin{eqnarray}x &=& 3+y \\ \nonumber x &=& 3 + \left(-\frac{5}{2}\right) \\ \nonumber x &=& \frac{6-5}{2} \\\nonumber x &=& \frac{1}{2}\end{eqnarray}$$
• Example 8 - Determine the value of \(x\) variable from the system of equations \(\begin{cases}2x+y&= 32 \\ -2x+3y &= 40\end{cases}\)
  Solution - To determine the value of \(x\) variable the value of \(y\) variable must be determine first. The first step is to sum both equations. By doing so, the x will be omitted from the equation and the value of \(y\) varaible will be determined. $$\begin{eqnarray}\begin{cases}2x+y&=& 32 \\ -2x+3y &=& 40\end{cases}\Bigg\}+\\ \nonumber 4y &=& 72/:4 \\ \nonumber y &=& 18\end{eqnarray}$$ Now that the value of \(y\) variable is determine it can be substituted into frist equation to obtain the value of \(x\) variable. $$\begin{eqnarray}2x + 18 &=& 32 \\\nonumber 2x &=& 14 \\ \nonumber x &=& 7\end{eqnarray}$$
• Example 9 - Determine the value of \(y\) variable from the system of equations \(\begin{cases}3x+4y+5&=0 \\ 7x-8y+16 &= 0\end{cases}\)
  Solution $$ \begin{eqnarray}\begin{cases}3x+4y+5&=&0/\cdot 2 \\ 7x-8y+16 &=& 0\end{cases}\\ \nonumber \begin{cases}6x+8y+10&=&0\\ 7x-8y+16 &=& 0\end{cases}\Bigg\}+\\\nonumber 13x+26 &=& 0 \\\nonumber 13x &=& -26/:13\\\nonumber x &=& -2\end{eqnarray}$$ $$\begin{eqnarray}3\left(-2\right) + 4y + 5 &=& 0 \\\nonumber 4y &=& 1 \\ \nonumber y &=& \frac{1}{4}\end{eqnarray}$$
• Example 10 - For variables \(a\) and \(b\) holds \(\begin{cases} a:b &= 3:4 \\a+b &= 21\end{cases}\)
  Solution $$\begin{eqnarray}\begin{cases} a:b &= 3:4 \\a+b &= 21\end{cases}\end{eqnarray}$$ From first equation express a in terms of b. $$\begin{eqnarray}\frac{a}{b} &=& \frac{3}{4}\\\nonumber 4a &=& 3b/:4 \\ \nonumber a &=& \frac{3}{4}b\end{eqnarray}$$ This expression will be inserted into second equation. $$\begin{eqnarray}a+b &=& 21 \\\nonumber \frac{3}{4}b + b &=& 21/\cdot 4 \\ \nonumber 3b+4b &=& 84 \\\nonumber 7b &=& 84/:7 \\\nonumber b &=& 12 \end{eqnarray}$$ Now that the value of variable b is obtained the value of variable a can be determined by inserting the value of variable b into the expression for determining variable a. $$\begin{eqnarray}a &=& \frac{3}{4}\cdot 12 \\ \nonumber a &=& 9.\end{eqnarray}$$
• Example 11 - If \(x^2 - y^2 = 75\) and \(x+y = 15\) determine how much it is \(x-y\) and 2x-2y+1 ?
  Solution From second equation express x over the variable y and insert it into first equation. $$\begin{eqnarray}x + y &=& 15 \\ \nonumber x&=& 15-y\end{eqnarray}$$ Now insert previosly obtained expression into the first equation. $$ \begin{eqnarray}x^2-y^2&=& 75\\\nonumber (15-y)^2 -y^2 &=& 75 \\\nonumber 225 - 30y + y^2 -y^2 &=& 75 \\\nonumber -30y &=& -150\\\nonumber y &=& 5\end{eqnarray}$$ Now that the value of y is determine we can determine the value x. $$\begin{eqnarray}x &=& 15-y \\\nonumber x&=& 15-5 \\\nonumber x&=& 10\end{eqnarray}$$ The value of the expression \(x-y\) can be determine by inserting value of x and y. $$ x-y = 10 - 5 = 5.$$ The value of the second expression \(2x-2y+1\) is determined in the same way. $$2x-2y+1 = 2\cdot 10 - 2\cdot 5 + 1 = 20-10+1 = 11.$$
• Example 12 - The number a is greather than positive number b for the value of 3. The ratio of a and b is equal to 5:3. Then a is equal to?
  Solution First write the system of equations. $$\begin{cases}a &= 3+b\\ \frac{a}{b} &= \frac{5}{3}\end{cases}$$ Insert first equation into te second equation. $$\begin{eqnarray}\frac{a}{b} &=& \frac{5}{3}\\ \nonumber \frac{3+b}{b}&=& \frac{5}{3}/\cdot 3b\\\nonumber9+3b &=& 5b\\\nonumber b&=&\frac{9}{2}\end{eqnarray}$$ Insert the variable b into first equation. $$\begin{eqnarray}a&=&3+b \\ \nonumber a&=& 3+\frac{9}{2}\\ \nonumber a&=&\frac{15}{2}\end{eqnarray}$$
• Example 13 - In some enclosed space we have bunnies and ducks. If we know that in this enclosed space there are total of 35 heads and 94 legs determine how much number of bunnies in the enclosed space exists ?
  Solution Both bunny and duck have one head, each. However, the bunny has 4 legs while ducks have only 2 legs. With first statement we can create equation $$\begin{eqnarray}x+y &=& 35\\ \nonumber x &=& 35-y\end{eqnarray}$$ With second statement the following exquation is obtained. $$4x + 2y = 94 $$. Now we can insert first equation into the second equation. $$\begin{eqnarray}4x+2y &=& 84 \\ \nonumber 4(35-y) +2y &=& 94 \\\nonumber -2y &=& -46 \\\nonumber y &=& 23\end{eqnarray}$$ This means that 23 ducks lives in enclosed spce. To calculate the number of bunnies simply insert the number of ducks into the first equation.$$ x = 35 - y = 35-23 = 12. $$ So there are total of 12 bunnies living in the encolsed space.