# solve the system: 2x-5y= 3 x+ 4y = 8

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2x-5y = 3...........(1)

x+ 4y = 8............(2)

We will use the substitution method to solve.

Let us rewrite eq. (2):

x+ 4y = 8

==> x= 8-4y

Now substitute x value in eq. (1):

2x-5y = 3

==> 2(8-4y) - 5y = 3

Open brackets:

==> 16 - 8y - 5y = 3

==> -13y = -13

**==> y= 1**

but x= 8- 4y

==> x= 8- 4*1 = 4

**==> x= 4**

First, we can say that x = 8 - 4y. Then we can substitute that into the first equation.

Once we have done that, we can make the following new equation:

2 (8 - 4y) - 5y = 3

This becomes

16 - 8y - 5y = 3

By subtracting 16 from both sides, we get

-8y - 5y = -13

or

-13 y = -13

By dividing both sides by -13, we get

y = 1

Now we substitute that back into the first equation and we get

x + 4 = 8

or

x = 4

2x - 5y=3 -------1 x+4y=8

x + 4y = 8 -------2

multiply equation 2 by 2 gives 2(x + 4y) = 8 * 2 2x + 8y = 16

2x - 5y = 3 -------1

2x + 8y = 16 -----3 Subtract 3 from 1 ngives

-5y - 8y = 3 - 16

-13y = -13

y=1

x + 4y = 8

x = 8 - 4y

x = 8 - 4(1)

x = 8 - 4

x = 4

Therefore (x,y) =( 4,1)

2x-5y = 3......(1)

x+4y = 8........(2)

We use the substitution method.

From (2) , we have x = 8-4y. We substitute x = 8-4y in (1):

2(8-4y) - 5y = 3

16 -8y -5y = 3

16 -13y = 3

16-3 = 13y

13 = 13y

13/13 = y

y = 1.

Substitute y = 1 in (1):

2x-5*1 = 3

2x-5 = 3

2x= 3+5 = 8

x = 8/2 = 4

x = 4 and y = 1.

We'll note the equations of the system as:

2x-5y= 3 (1)

x+ 4y = 8 (2)

We'll solve the system using elimination method. For this reason, we'll multiply (2) by -2:

-2x -8y = -16 (3)

Now, we'll add (3) to (1):

2x-5y-2x -8y=3-16

We'll combine and eliminate like terms:

-13y = -13

We'll divide by -13 and we'll get:

**y = 1**

We'll substitute y in (1):

2x-5y= 3

2x = 3+5

2x = 8

We'll divide by 2:

**x = 4**

**The solution of the system is: {(4,1)}.**