# solve for x , y2x - 3y = -1 x+5y =7

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

x+5y = 7........(2)

Using the substituting method:

Rewrite (2);

x= 7- 5y

Now substitute n (1)

2x-3y = -1

2(7-5y) - 3y = -1

14 - 10y -2y = -1

12y = 15

==> y = 15/12 = 5/4

==> **y= 5/4**

==> x= 7-5y = 7-5(5/4)

= 7- 25/4 = 28-25/4 = 3/4

==> **x= 3/4**

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

x + 5y = 7 ... (2)

Multiplying equation (2) by 2:

2x + 10y = 14 ... (3)

Subtracting equation (1) from equation (3):

2x - 2x + 10y + 3y = 14 + 1

==> 13y = 15

Therefore:

y = 15/13

Substituting this value of y in equation (2):

x + 5*(15/13) = 7

==> x + 75/13 = 7

==> x = 7 - 75/13 = 16/13

Answer:

x = 16/13,

y = 15/13

Because the equations of the system are linear, we'll use the matrix formed by the coefficients of the variables to calculate the determinant of the system:

2 -3

det A =

1 5

det A = 10+3 = 13

det A = 13 different from zero.

Now, we'll calculate x:

x = det x/ det A

-1 -3

det x =

7 5

det x = -5 - 21 = -26

x = -26/13

**x = -2**

Now, we'll calculate y:

2 -1

det y =

1 7

det y = 14-1

det y =15

y = det y/ det A

y = 15/13

**y = 15/13**

**The solution of the system is: {(-2 , 15/13)}**

Given 2x-3y=-1 and x+5y=7 we have to find x and y.

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

x+5y=7 ...(2)

Multiply (2) by 2 and subtract (1). We get

2x+10y-2x+3y=14-(-1)

=>13y=15

=>y=15/13

Substituting y=15/13 in (2) we get

x+5*(15/13)=7 or x=7-45/13 or x=(91-75)/13=16/13

**Therefore x=16/13 and y=15/13**

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

x+5y = 7.......(2)

Solution:

We can eliminate x by Eq(1) -2*(2) and solve for y:

(2x-3y) -2(x+5y) = -1-2*7

-3y-10y = -1-14 = -15

-13y =-15.

y = -15/-13 = 15/13

We can eliminate y by the operation 5*Eq(1) +3*Eq(2):

5(2x-3y)+3(x+5y) = 5*(-1)+3*7 = 16

10x+3x =16

13x=16

13x/13 = 16/13

x = 16/13

So (x,y) = (16/13 , 15/13).