# Solve the system: 2x+y=9 10x-4y=36

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We have to solve for x and y using the system

2x + y = 9...(1)

10x - 4y = 36...(2)

Using (1)

2x + y = 9

=> y = 9 - 2x

substitute this in (2)

=> 10x - 4( 9 - 2x ) = 36

=> 10x - 36 + 8x = 36

=> 18x = 72

=> x = 4

y = 9 - 2*4 = 9 - 8 = 1

**Therefore x = 4 and y = 1.**

This is a simple quadratic function.

To solve x or y, we need to put the coefficient of x or y same for each equation so that x or y can be eliminated when we subtract or add the two equations.

Let's eliminate y.

The coefficient of y in the first equation is 1, and in the second equation, it is -4. Therefore, if we multiply 4 to the first equation, the coefficients of ys in the both equation will be same.

4(2x +y=9)

==> 8x +4y= 36

Now, let's add both the equation.

8x+4y=36

+)10x-4y=36

18x = 72

18x=72

x=72/18 = 4

Now, let's solve y.

Substitute x for 4. Then, the first equation is

2(4) + y=9

8+y=9

y=1

Therefore, x=4, y=1.

Let's check if these two values are correct.

If we substitute these values in the second equation,

10(4) -4(1)=36

40-4=36

Since the equation is correct, x=4 and y=1.

Solve the system:

2x+y=9........(1).

10x-4y=36...(2).

(1)*4+(2) gives (2x+y)4+(10x-4y) = 9*4+36 = 72.

18x = 72.

x= 72/18 = 4.

(1)*5 - (2) gives (2x+y)5 - (10x-4y) = 9*5-36 = 9.

9y = 9.

y = 9/9 = 1.

x = 4 and y = 1.

Before solving the system, we'll perform some changes into the given equations.

We'll change the first equation.

-2x-y=-9

We'll add 2x:

-y = 2x - 9

We'll multiply by -1:

y = -2x + 9 (1)

Now, we'll factorize the second equation by 2:

2(5x-2y)=36

We'll divide by 2 both sides:

5x-2y=18 (2)

We'll solve the system using substitution.

We'll substitute (1) in (2):

5x - 2(-2x + 9) = 18

We'll remove the brackets:

5x + 4x - 18 = 18

We'll combine like terms:

9x = 18 + 18

x = 2*18/9

x = 4

We'll substitute x = 4 in (1):

y = -2*4 + 9

y = 9 - 8

y = 1

**The solution of the system is: (4 ; 1).**