There are two numbers X and Y. We know that X+Y=36. Let us say that X is the smaller number: 4X+15=Y+4. We will first solve for Y in terms of X: 4X+11=Y. Now we will substitute this back into the first equation: X+4X+11=36 , 5X=25, X=5. This makes Y=31. To check this:4*5+15=35, 31+4=35.

you have two variables (unknowns) so you need two equations (for 3 varaibles, you need 3 equations, etc)that you can then add together or substitute into each other. start with let statements: let x be the smaller number, let y be the bigger number write your equations (words like sum and more mean addition, less and difference mean subtraction, product and times mean multiplication, quotient means division): 1) y+x=36 2) 4x+15=y+4: when simpified and rearranged is y-4x=11 remember when rearrabging equations to BS (Both sides! whatever you do to one side, do to the other) option 1:substitute y=36-x (equation 1 rearranged) (36-x)-4x=11 - now you have 1 equation with 1 variable: >simplify the equation, isolate the varaible, solve -5x=-25 x=5 : substitute this number back into an original equation, and check your answers option 2:add or subtract the equations so one variable cancels out,somtimes you have to multiply equations (every part of the equation) y+x=36 y-4x=11 (y-y)+(x-(-4x))=(36-11)>notice subtracting the -ve # 5x=25 > solve, and substitue back into an equation to solve for the other variable, check your answer These steps should work for any problem you have- carefully read problem, let statements, write equations, simplify and rearrange, substitute or add, substitute known varaible back into an equation, check your answers.

Let's call the larger number L and the smaller number S. S+L=36. 4S+15=4+L. Now subtract 4 from both sides of the second equation. Then we get 4S+11=L. Since we now know what L is in terms of S, let's plug it in to the first equation. So we get S+4S+11=36 or 5S+11=36. Subtract 11 from both sides and you get 5S=25. Divide both sides by 5 and you get S=5. Well, now that we know the smaller number is five and 4S+11=L then 4*5+11=L or 20+11=L or 31=L. So the small number is five and the large number is 31. To do these kind of problems just keep doing something to both sides of an equation to make it simple as possible and plug in one number that you already know something about into the other equation.