Question 22321
 SOLUTION
 THIS IS A SIMULTANEOUS EQUATION. 
  YOU CAN EITHER SOLVE IT BY ELIMINATION METHOD OR SUBSTITUTION METHOD.
 ELIMINATION METHOD HAS BEEN THE EASIEST AND FAST METHOD OF SOLVING ANY SIMULTANEOUS EQUATION.

   LET'S SEE:BY ELIMINATION METHOD.
         3X+2Y=4 --eq 1
         4X-5Y=4 --eq 2

 STEP 1
 MULTIPLY EQUATION ONE BY 4 THROUGHOUT AND SIMULTANEOUSLY MULTIPLY EQUATION TWO BY 3 THROUGHOUT.( 3 WAS DERIVE FROM CO-EFFICIENT OF X IN EQUATION ONE AND 4 WAS DERIVE FROM CO-EFFICIENT OF X IN EQUATION TWO,THIS IS TRUE ALL THE TIME FOR ANY SIMULATENEOUS EQUATION USING ELIMINATION METHOD APPROACH.      
 THOSE EQUATION WILL BECOME
         12X+8Y=16 --EQ 3
         12X-15Y=12--EQ 4
STEP TWO- SUBTRACT EQUATION 4 FROM EQUATION THREE
              23Y=4
                Y=4/23

TO GET THE VALUE OF X.
          PLUG IN  4/23 FOR Y ,IN ANY OF THE EQUATION TO GET X.
             3X+2Y=4 -EQ 1
             3X+2(4/23]=4
             3x+8/23=4
             3x=4-(8/23)(by collecting like terms}
             3x=(92-8}/23
             3x=84/23
              x={84/23}X (1/3}
              X=28/23

we can always check if we are right by proof evaluattion.
        by plugging both values of x and y in any of the equation respectively if we do so the left hand side must eqaul to the right hand side to certify that we are right.

  let'see- 
        by taking any equation,say equation one.
                        3x+2Y=4
                        3{28/23)+2(4/23)=4
                          (84/23)+(8/23}=4
                           3.65  + 0.35 =4
This tells us our value of x and y are correct,this  proof is true at all time to proof or check any given value of x and y,in any simulataneous equation problems.
Thank you.

 Moruph Osuolale
mosuolal@uncc.edu