So for this, I will be using the elimination method. So firstly, multiply both sides of the first equation by 2 and both sides of the second equation by 3:
![\left \{ {{8x-6y=52} \atop {9x+6y=33}} \right.](https://tex.z-dn.net/?f=%20%5Cleft%20%5C%7B%20%7B%7B8x-6y%3D52%7D%20%5Catop%20%7B9x%2B6y%3D33%7D%7D%20%5Cright.%20)
Now add the two equations to get
. From here you can solve for x. For this, just divide both sides by 17, and your first answer will be ![x=5](https://tex.z-dn.net/?f=%20x%3D5%20)
Now that we have the value of x, substitute it into either equation to solve for y:
![4*5-3y=26\\ 20-3y=26\\ -3y=6\\ y=-2\\ \\ 3*5+2y=11\\ 15+2y=11\\ 2y=-4\\ y=-2](https://tex.z-dn.net/?f=%204%2A5-3y%3D26%5C%5C%2020-3y%3D26%5C%5C%20-3y%3D6%5C%5C%20y%3D-2%5C%5C%20%5C%5C%203%2A5%2B2y%3D11%5C%5C%2015%2B2y%3D11%5C%5C%202y%3D-4%5C%5C%20y%3D-2%20)
<h3><u>In short, x = 5 and y = -2.</u></h3>
Check the picture below.
I'd like to point out, is 1:5 from R to T, since that matters, that way we know the ratio from RS is the 1 and the ratio ST is the 5.
![\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ R(-14,-1)\qquad T(4,-13)\qquad \qquad 1:5 \\\\\\ \cfrac{RS}{ST} = \cfrac{1}{5}\implies \cfrac{R}{T} = \cfrac{1}{5}\implies 5R=1T \implies 5(-14,-1)=1(4,-13)\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft.%20%5Cqquad%20%20%5Cright.%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%0A%5C%5C%5C%5C%5C%5C%0AR%28-14%2C-1%29%5Cqquad%20T%284%2C-13%29%5Cqquad%20%5Cqquad%201%3A5%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BRS%7D%7BST%7D%20%3D%20%5Ccfrac%7B1%7D%7B5%7D%5Cimplies%20%5Ccfrac%7BR%7D%7BT%7D%20%3D%20%5Ccfrac%7B1%7D%7B5%7D%5Cimplies%205R%3D1T%0A%5Cimplies%20%0A5%28-14%2C-1%29%3D1%284%2C-13%29%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
Width = (7,6) - (7,3) = 3
height (7,3) - (2,3) = 5
Its dont know .............