4.2.4 Exercises

1. For example, solving the first equation ($3x+y=4$) for $y$ results in $y=4-3x$. Then this equation can be substituted into the second equation ($-x+2y=1$): $-x+2(4-3x)=1\iff -x+8-6x=1\iff -7x=-7\iff x=1$. With this value of $x$, for example, the second equation results in $3·1+y=4\iff y=1$. So, the solution set is $L=\{(1;1)\}$.
   Of course, one could start differently: For example, one could solve the first equation for $x$ and substitute the solution for $x$ into the second equation, to get a solution for $y$; or one could start generally with the second equation and solve this in the first step for $x$ or for $y$. Hence, there are some details in the approach.
   
2. Solving, for example, the first equation ($-x+4y=5$) for $x$ results in $x=4y-5$. Then, this equation can be substituted into the second equation ($2x-8y=-10$): $2(4y-5)-8y=-10\iff 8y-10-8y=-10\iff 0=0$. So, this is not a new statement, in other words: the second equation does not contain new information. Thus, in this case the solution set $L$ contains an infinite number of solution pairs $(x;y)$ which can be parametrised by a real number $t$. If one chooses, for example, $y=t$, the solution set reads $L=\{(4t-5;t)\hspace{0.5em}:\hspace{0.5em}t\in ℝ\}$. This solution set can be visualised as a line in two-dimensional space:
   
   Accordingly, other parametrisations of the solution set are possible, for example, by choosing $x\in ℝ$ as free parameter and characterising the above line by its slope and its $y$-intercept, i.e.  $L=\{(x;\frac{1}{4}x+\frac{5}{4})\hspace{0.5em}:\hspace{0.5em}x\in ℝ\}$.
   

1. Multiplying the second equation ($x+2y=3$) by $(-2)$ results in equation $(2\text{'})$: $-2x-4y=-6$. Then the last equation is added to the first equation ($2x+4y=1$): $2x+4y-2x-4y=1-6\iff 0=-5$. This is a contradiction! Hence, the solution set for this system of equations is the empty set $L=\text{∅}$.
   
2. Multiplying the first equation ($-7x+11y=40$) by $2$ results in the equation $(1\text{'})$: $-14x+22y=80$. Multiplying the second equation by $7$ results in the equation $(2\text{'})$: $14x+35y=91$. Subsequently, adding equation $(1\text{'})$ and equation $(2\text{'})$ results in $-14x+22y+14x+35y=80+91\iff 57y=171\iff y=3$. Substituting this value of $y$, for example, in the second equation results in $2x+5·3=13\iff 2x=13-15\iff 2x=-2\iff x=-1$. Hence, the solution set $L$ reads $L=\{(-1;3)\}$.
   

The first equation ($2x=2$) is equivalent to $x=1$: This equation describes a line parallel to the $y$-axis through the point $(1;0)$ on the $x$-axis. The second equation ($x+3y=4$) can be transformed into $y=-\frac{1}{3}x+\frac{4}{3}$. This equation also describes a line, this time its slope is $-\frac{1}{3}$ and its $y$-intercept is $\frac{4}{3}$. The graphs are as in the following figure. >From this figure one reads off the coordinates of the intersection point as $(x=1;y=1)$. Hence, the solution set is $L=\{(1;1)\}$.