# x/2 + 3y =5 x+3y =3 slove for x and y

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x/2 +3y =5...(1)

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

We are going to use the elimination method:

First we will subtract (1) from (2)

==> x/2 = -2

Multiply by 2

==> x= -4

Now to calculate y, we are going to use either (1) or (2) and substitute with x= -4

In (1) :

x/2 + 3y =5

-4/2 + 3y = 5

-2+ 3y =5

Add 2 to both sides:

==> 3y = 7

Now divide by 3,

==> y= 7/3

In the first equation, we notice that we have the common denominator 2, so, we'll multiply 3y and 5, by 2:

x + 2*3y = 2*5

x + 6y = 10

Now, we'll add -6y both sides:

x = -6y + 10 (1)

We'll write the second equation:

x + 3y = 3

We'll add -3y both sides:

x = -3y + 3 (2)

From (1) and (2), we'll get:

-6y + 10 = -3y + 3

We'll isolate y to the left side:

-6y + 3y = 3 - 10

-3y = -7

We'll divide by -3:

y = 7/3

We'll substitute y by it's value, into the relation (1):

x = -6*(7/3) + 10

x = -14 + 10

x = -4

The solution of the system is: {(-4 , 7/3)}

To solve x/2 + 3y =5 and x+3y =3.

Solution:

We have two linear simultaneous equations in two variables.We can use substitution method or elimination method .

If we subtract the equations the terms 3y cancels and the result would be an equation in one variable x and we can solve for x.

But we use the method of substitution. From the second equation we get x = 3-3y. We make substitution of 3-3y for x in 1st equation and get:

(3-3y)/2 +3y = 5. Multiplying by 2, we get:

3-3y +6y = 10.

3+3y = 10.

3y =10-3 =7.

y = 7/3.

Substituting y=7/3 in first equation, we get: x/2 + 3(7/3) = 5. Or

x/2 +7 =5. Or

x +14 = 10. Or

x = 10-14 = -4.

So x=-4 and y = 7/3.

Given:

x/2 + 3y = 5 ... (1)

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

Subtracting equation (2) from equation (1) we get:

x/2 - x + 3y - 3y = 5 - 3

-x/2 = 2

x = 2*(-2) = -4

Substituting this value of x in equation (2) we get:

-4 + 3y = 3

3 y = 3 +4 = 7

Therefore:

y = 7/3

Answer:

x = -4, y = 7/3