# Please help me! Thanks in advance! The follow problem falls under Algebra 2. It is Systems of Equations. 1. 5x + 4y = 6 -2x - 3y = -1

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`5x+4y = 6 ---(1)`

`-2x-3y = -1 ----(2)`

By multiplying (1) and (2) with some constant values, we should make same coefficient in x or y.

`(1)xx2`

`10x+8y = 12 ----(1)'`

`(2)xx5`

`-10x-15y = -5-------(2)'`

Now if you see carefully there are two terms with 10x and -10x in (1)' and (2)' respectively. So we add up those two we can cancel out x from both equations.

`(1)'+(2)'`

`10x+8y+(-10x-15y) = 12+(-5)`

`10x+8y-10x-15y = 12-5`

`-7y = 7`

`y = -1`

By substituting the value of y in (1) we can get the value of x.

`5x+4y = 6`

`5x+4(-1) = 6`

`5x-4 = 6`

`5x = 6+4`

`5x = 10`

`x = 2`

*So the answers are x = 2 and y = -1*

`5x+4y=6...................(i)`

`-2x-3y=-1` (ii)

multiply (i) by 3 and (ii) by 4,add the resultng equations

`15x-8x=18-4`

`7x=14`

`x=14/7=2`

Multiply (i) by 2 and (ii) by (5), add the equations

`8y-15y=12-5`

`-7y=y`

`y=(7)/(-7)=1`

Thus solution of equations is x=2 ,y=-1

THe point of intersection (2,-1) of lines is solution of the system of equations.