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
[tex]x\text{ = }\frac{4\text{ -2y}}{7}[/tex]In equation 2
-5x -3y = 5
-5x = 5 +3y
[tex]x\text{ = }\frac{5+3y}{-5}\text{ = -}\frac{(5\text{ +3y)}}{5}[/tex]Thus , the x in equation 1 is equvalent to x in equation 2
Hence, we will equate their values
[tex]\begin{gathered} \frac{4-2y}{7}\text{ = -}\frac{(5+3y)}{5} \\ \text{cross multiply} \\ 5\text{ }\times\text{ (4-2y) = 7 }\times(-5-3y) \\ \text{removing bracket} \\ 5\text{ }\times4\text{ - 5}\times2y\text{ = 7}\times-5\text{ -7}\times3y \end{gathered}[/tex][tex]\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} \\ \frac{11y}{11}=\frac{-55}{11} \\ y\text{ = -5} \end{gathered}[/tex]substitute y = -5 in equation 1 to obtain x
7x +2y = 4
[tex]\begin{gathered} 7x\text{ + 2 (-5) = 4} \\ 7x\text{ - 10 = 4} \\ 7x\text{ = 4+10} \\ 7x\text{ =14} \\ \text{divide both side by 7} \\ \frac{7x}{7}=\frac{14}{7} \\ x\text{ = 2} \end{gathered}[/tex]The solutions are x= 2 and y = -5