Question

Solve the system of equations 5x-2y=88 3x+4y=58 show all work

Answer 1

`\left\{\begin{array}{ccc}5x-2y=88&1^o\\3x+4y=58&2^o\end{array}\right\\\\1^o\ 5x-2y=88\ \ \ \ |\text{subtract 5x from both sides}\\-2y=-5x+88\ \ \ |\text{divide both sides by (-2)}\\y=2.5x-44\\\\\text{substitute the value of y to the equation}\ 2^o`


`3x+4\cdot(2.5x-44)=58\\3x+10x-176=58\\13x-176=58\ \ \ |\text{add 176 to both sides}\\13x=234\ \ \ \ |\text{divide both sides by 13}\\x=18\\\\\text{substitute value of x to the equation}\ 1^o\\\\y=2.5\cdot18-44=45-44=1\\\\Answer:\ \boxed{\left\{\begin{array}{ccc}x=18\\y=1\end{array}\right}`

Answer 2

The given equations are:

5x - 2y = 88
3x + 4y = 58

Multiplying the 1st equation by 2, we get the new set of equations as:

10x - 4y = 176
3x + 4y = 58

Adding the two equations, we get:

10x - 4y + 3x + 4y = 176 + 58
13x =234
x = 18

Using the value of x in 1st equation, we get:

5(18) - 2y = 88
- 2y = 88 -5(18)
-2y = -2
y = 1

So, the solution of the equation is (18, 1)