# What is the quickest way to find the sum of $x$ and $y$ in this system of equations?

\begin{align*} 2\left( x-\frac{1}{3}\right) -\frac{3}{2}\left( y-\frac{1}{6}\right)&=0 \tag 1\\ 3\left( y-\frac{1}{2}\right) +\frac{8}{3}\left( x-\frac{1}{6}\right)&=0 \tag 2\\ \end{align*} Consider the system of equations above. If $(x,y)$ is the solution to the system, then what is the value of the sum of $x$ and $y$?

I've used the solving for $x$, and then, finding $y$ method, but I feel, that there must be a faster way to find $x+y$. Does anyone know any tricks?

• Solving the system consists in reducing its matrix to a triangular one and performing back substitution. However, you can aim instead to getting a row $[1,1]$ in the matrix by doing row transformations. If you do that the right hand side is the value of $x+y$ – Hellen Jul 21 '17 at 1:11
• Another way can be to put $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$ and solve only for $a$. – Hellen Jul 21 '17 at 1:19

\begin{align*} 2\left( x-\frac{1}{3}\right) -\frac{3}{2}\left( y-\frac{1}{6}\right)&=0 \\ 3\left( y-\frac{1}{2}\right) +\frac{8}{3}\left( x-\frac{1}{6}\right)&=0 \\ \end{align*} Multiply the first by $2$ and the second by $21$ \begin{align*} 4 x-3 y&=\frac{5}{6} \\ 56 x+63 y&= \frac{245}{6}\\ \end{align*} Then add the two equations $$60x+60y= \frac{250}{6}\rightarrow x+y=\frac{25}{36}$$

In general, we must multiply $(1)$ by $a$ and $(2)$ by $b$ so that when the two equations are added we get the coefficients of $x$ and $y$ the same: $$2a+\frac{8}{3}b=3b-\frac{3}{2}a \Rightarrow 21a=2b \Rightarrow a=2, b=21.$$

Hence: $$60(x+y)=(\frac23-\frac14)\cdot2+(\frac32+\frac49)\cdot21 \Rightarrow x+y=\frac{25}{36}.$$

• +1, as far as I can se this is the only method that does not calculte x and y before x+y and seems to be shorter than other methods to calculate x+y, but this is hard to judge. Nevertheless the savings are so small that I think one should calculate x and y with the standard methods and then calculate x+y. – miracle173 Jul 22 '17 at 6:53

Begin with the system of two equations: \begin{align*} 2\left( x-\frac{1}{3}\right) -\frac{3}{2}\left( y-\frac{1}{6}\right)&=0, \\ 3\left( y-\frac{1}{2}\right) +\frac{8}{3}\left( x-\frac{1}{6}\right)&=0. \\ \end{align*} Multiply through the first equation by $\color{blue}{12}$ and multiply through the second equation by $\color{green}{18}$ so that we can clear the denominator for both equations: \begin{align*} \color{blue}{4}\cdot 2\left( \color{blue}{3}\left( x-\frac{1}{3}\right)\right) - \color{blue}{2}\cdot \frac{3}{2}\left(\color{blue}{6}\left( y-\frac{1}{6}\right) \right)&=0, \\ \color{green}{9}\cdot 3\left( \color{green}{2}\left( y-\frac{1}{2}\right)\right) +\color{green}{3}\cdot \frac{8}{3}\left( \color{green}{6}\left(x-\frac{1}{6}\right) \right)&=0. \\ \end{align*} Now, multiply out the numbers in blue and in green: \begin{align*} 8\left( 3 x-1 \right) - 3 \left(6y-1\right)&=0, \\ 27 \left( 2y-1 \right) + 8 \left( 6x-1 \right)&=0. \\ \end{align*} Expand: \begin{align*} 24x -8 -18y+3 &= 0, \\ 54 y - 27 + 48x-8 &= 0. \\ \end{align*} Simplify further as: \begin{align*} 24x - 18y &= 8-3=5, \\ 48x + 54 y &= 27+8=35. \\ \end{align*} Write the two equations in matrix form: $$\begin{pmatrix} 24 & -18 \\ 48 & 54 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 5 \\ 35 \\ \end{pmatrix}.$$ Write this as an augmented matrix: $$\begin{pmatrix} 24 & -18 & 5\\ 48 & 54 & 35 \\ \end{pmatrix}$$ and then row-reduce (multiply the first row by $-2$ and then add the corresponding numbers to the second row): $$\begin{pmatrix} 24 & -18 & 5\\ 0 & 90 & 25 \\ \end{pmatrix}.$$ Divide the second row by $5$ to get: $$\begin{pmatrix} 24 & -18 & 5\\ 0 & 18 & 5 \\ \end{pmatrix}.$$ Add the second row to the first row: $$\begin{pmatrix} 24 & 0 & 10\\ 0 & 18 & 5 \\ \end{pmatrix}.$$ Divide the first row by $24$ and divide the second row by $18$: $$\begin{pmatrix} 1 & 0 & \frac{5}{12}\\ 0 & 1 & \frac{5}{18} \\ \end{pmatrix}.$$ So $$x= \frac{5}{12} \mbox{ and } y= \frac{5}{18},$$ and we conclude $$\boxed{x+y = \frac{5}{12} + \frac{5}{18} = \frac{25}{36}}.$$

Hint:  let $s=x+y\,$, then substituting $x=s-y$ back into the equations gives:

\begin{cases} \begin{align} 2\left((s-y)-\frac{1}{3} \right) - \frac{3}{2}\left(y-\frac{1}{6}\right) &= 0 \quad\iff\quad 2s - \frac{7}{2}y = \frac{5}{12}\\ \frac{8}{3}\left((s-y)-\frac{1}{6}\right) + 3\left(y-\frac{1}{2}\right) &= 0 \quad\iff\quad \frac{8}{3}s + \frac{1}{3}y=\frac{35}{18} \end{align} \end{cases}

Eliminate $y$ between the latter equations (for example multiply the first one by $1/3\,$, the second one by $7/2$ and add up), then solve the resulting equation for $s=x+y$.

• Nice! It is faster to eliminate the $y$ and thus solve for $s$ using the last two equations. – Mee Seong Im Jul 21 '17 at 4:11