Solve each question algebraically: either using the elimination or the substitution method.

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

multiply the bottom equation by 2 and add the equations.

3 2/3 x=11
then solve for x. I get 3

THANKS!

To solve the given system of equations algebraically using the elimination or substitution method, we will eliminate one variable by multiplying one or both equations by appropriate constants to make the coefficients of one of the variables the same. Then we will either add or subtract the two equations to eliminate that variable and solve for the remaining variable.

Let's solve the system using the elimination method:

Step 1: Multiply both sides of both equations by the least common multiple (LCM) of the denominators to clear the fractions. In this case, the LCM of 2 and 4 is 4. Multiply through on both equations by 4 to get rid of the fractions:

Equation 1: 3x - 1/2y = 5 => 12x - 2y = 20
Equation 2: 1/3x + 1/4y = 3 => 4/3x + 1y = 12

Step 2: Now we have two equations with similar coefficients for the 'y' term. Add the equations together to eliminate the variable 'y'.

(12x - 2y) + (4/3x + 1y) = 20 + 12
12x + (4/3)x - 2y + 1y = 32
(12 + 4/3)x - y = 32
(36/3 + 4/3)x - y = 32
40/3x - y = 32

Step 3: Simplify the equation:
40/3x - y = 32

Now we have one equation with one variable. Let's isolate 'x':

40/3x - y = 32
40/3x = y + 32
40x = 3(y + 32)
40x = 3y + 96
x = (3y + 96)/40
x = (3/40)y + 24/5

So, the solution is x = (3/40)y + 24/5.

Now, let's substitute the value of x back into one of the original equations to solve for 'y':

Using Equation 1: 3x - (1/2)y = 5
3((3/40)y + 24/5) - (1/2)y = 5
(9/40)y + 72/5 - (1/2)y = 5
(18/80)y + (144/80)y = 5 - 72/5
(18/80 + 144/80)y = (25/5 - 72/5)
(162/80)y = (-47/5)
y = (-47/5)*(80/162)
y = -3760/810
y = -188/81

So, the solution to the system of equations is x = (3/40)y + 24/5 and y = -188/81.