We have two equations to solve simultenously
7x + 2y = 4-------------1
-5x - 3y = 5-------------2
make x subject of the formular in both equation 1 and 2
In equation 1
7x + 2y = 4
7x = 4 - 2y
![x\text{ = }\frac{4\text{ -2y}}{7}](https://img.qammunity.org/2023/formulas/mathematics/college/lk6zux4c24bkvn4jtgbxoq9p3uoto3799x.png)
In equation 2
-5x -3y = 5
-5x = 5 +3y
![x\text{ = }(5+3y)/(-5)\text{ = -}\frac{(5\text{ +3y)}}{5}](https://img.qammunity.org/2023/formulas/mathematics/college/nu889z2vlfr2kgc4kzwnttqed1yu1cvzwm.png)
Thus , the x in equation 1 is equvalent to x in equation 2
Hence, we will equate their values
![\begin{gathered} (4-2y)/(7)\text{ = -}((5+3y))/(5) \\ \text{cross multiply} \\ 5\text{ }*\text{ (4-2y) = 7 }*(-5-3y) \\ \text{removing bracket} \\ 5\text{ }*4\text{ - 5}*2y\text{ = 7}*-5\text{ -7}*3y \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/m8pgsjphgxiaviwkur31r9hd711ezctlns.png)
![\begin{gathered} \text{ 20 -10y = -35 -21y} \\ \text{collect like terms} \\ -10y\text{ +21y = -35 -20} \\ 11y\text{ = -55} \\ \text{divide both side by 11} \\ (11y)/(11)=(-55)/(11) \\ y\text{ = -5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i7e6hhtx3wkyit5fjerpv84jkoxf09faai.png)
substitute y = -5 in equation 1 to obtain x
7x +2y = 4
![\begin{gathered} 7x\text{ + 2 (-5) = 4} \\ 7x\text{ - 10 = 4} \\ 7x\text{ = 4+10} \\ 7x\text{ =14} \\ \text{divide both side by 7} \\ (7x)/(7)=(14)/(7) \\ x\text{ = 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fx6xxpdkjgc22hb2rgnkla5w5cu6vj7pyf.png)
The solutions are x= 2 and y = -5