# Solve the system x+2y=3                            4x+5y=6

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

4x + 5y = 6........(2)

Let us use the substitution method:

from (1):

x = 3- 2y

Now substitute in (2):

4x + 5y = 6

4(3-2y) + 5y = 6

12 - 8y + 5y = 6

-3y = -6

y= 2

x = 3-2y = 3- 2*2 = -1

x= -1

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nisarg | Student

x+2y=3

4x+5y=6

multiply (x+2y=3) by 4 and subtract the equation 4x+5y=6

4x+8y=12

-  4x+5y=6

0x+3y=6

y=2

so plug in y in any equation to get x

x+4=3

x=-1

there you go :)

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jess1999 | Student

x   + 2y = 3

4x + 5y = 6

First multiply every single digit in the first equation by 4

By doing that, your equation should look like

4x + 8y = 12

4x + 5y = 6     now, subtract 4x with 4x. Which means also subtract 8y with 5y and 12 with 6

By subtracting, your equation should look like

3y = 6 now divide both sides by 3

By dividing, your equation should look like

Now, plug 2 into one of the equation

By doing that, you should get

x + 4 = 3 subtract both sides by 4

By subtracting, your equation should look like

So your answer to this solution is x = -1   ;   y = 2

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placid07 | Student

x+2y=3

4x+5y=6

ELIMINATION:

(-4)(x+2y=3)

4x+5y=6

-4x-8y=-12

0x-3y=-6

-3y=-6

y=2

Substitution:

4x+5y=6

y=2

4x+5(2)=6

4x+10=6

4x=-4

x=-1

SOLUTION SET: (-1,2)

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krishna-agrawala | Student

x + 2y = 3   ...   (1)

4x + 5y = 6   ...   (2)

Multiplying equation (1) by 4:

4x + 8y = 12   ...   (3)

Subtracting equation (2) from equation (3):

4x - 4x + 8y - 5y = 12 - 6

==> 3y = 6

Therefore:

y = 6/3 = 2

Substituting this value of y in equation (1):

x + 2*2 = 3

==> x = 3 - 4 = -1

x = -1

y = 2

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neela | Student

x+2y =3

4x+5y = 6.

Solution:

a1x+b1y =c1........(1)

a2x+b2y =c2.........(2)

Eliminate y by    (1)*b2 - (2)*b1

( a1b2-a2b1)x= b2c1- b1c2. Now solve for x:

x = ( b2c1-b1c2)/(a1b2-a2b1)

Similarly eliminate x by the operation (1)*a2- (2)*a1

(a2b1-a1b2)y = a2c1-a1c2

y = (a2c1-a1c2)/(a2b1-a1b2)

In this case  a1 = 1 , b1 =2 , c1 = 3, a2 = 4, b2 =5 , c2 =6.

So x=  ( b2c1-b1c2)/(a1b2-a2b1) = (5*3-2*6)/(1*5 - 4*2) = 3/-3 = -1.

y = (a2c1-a1c2)(a2b1-a1b2) = (4*3-1*6)/(4*2-1*5) = 6/3 =2.

Therefore (x , y) = (-1 ,2)

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thewriter | Student

We are given the equations

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

and 4x+5y=6                           ...(2)

Multiply (1) by 4 and subtract (2), we get

4x+8y-4x-5y=12-6

=>3y=6

=>y=2

Substituting y=2 in (1), x+4=3 or x=-1

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giorgiana1976 | Student

We'll solve the system using the substitution method.

We'll note the equations of the system as:

x+2y=3 (1)

4x+5y=6 (2)

We'll re-write (1):

x = -2y+3 (3)

We'll substitute x in (2):

4(-2y+3)+5y=6

We'll remove the brackets:

-8y + 12 + 5y - 6 = 0

We'll combine like terms:

-3y + 6 = 0

We'll subtract 6 both sides:

-3y = -6

We'll divide by -3:

y = 2

We'll substitute y in (3):

x = -2y+3

x = -2*2 + 3

x = -4+3

x = -1

The solution of the system is: {(-1 , 2)}.

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