x= students

y= adults

2x + 3y = 1050

x + y = 400

2x + 3y = 1050

– 2x + 2y = 800

= y = 250

x + 250 = 400

x = 150

there were 250 adults and 150 students

you get 3 equations: a million) 4x + 6y – 3z = 20 2) x – 5y + z = 15 3) -7x + y + 2z = a million and you have 3 variables. What i might do, is substitute between the variables in the different 2 equations, i’ll do this with z. by potential of doing this, you get to new equations to get x and y, that’s plenty much less complicated because it is straightforward. i’ll get the value of z in equation 2) so: 2) x – 5y + z = 15 subtract x -5y + z = 15 – x upload 5y z=15 – x + 5y Now I replace z in the different 2 equations. So: a million) 4x + 6y – 3z = 20 replace z 4x + 6y – 3(15 – x + 5y) = 20 so: 4x + 6y – 40 5 + 3x – 15y = 20 upload equivalent matters 7x – 9y – 40 5 = 20 upload 40 5 7x-9y=sixty 5 this may well be now equation 4) now I replace z in equation 3) so: 3) -7x + y + 2z = a million replace z -7x + y + 2(15 – x + 5y) = a million so: -7x + y + 30 – 2x + 10y = a million upload equivalent matters -9x + 11y + 30 = a million subtract 30 -9x + 11y = -29 this may well be equation 5) So, doing an equation device between equation 4 and 5 4) 7x-9y=sixty 5 5) -9x + 11y = -29 i’ll substitute the subject of the two equations to x equals to a minimum of something to get the value of y so: 4) 7x – 9y = sixty 5 upload 9y 7x=sixty 5+9y divide by potential of seven x=(sixty 5+9y)/7 5) -9x + 11y = -29 upload 9x 11y = -29 + 9x upload 29 11y + 29 = 9x divide by potential of 9 (11y + 29)/9=x for this reason: (11y + 29)/9=(sixty 5+9y)/7 multiply by potential of 9 11y + 29 = 9(sixty 5+9y)/7 multiply by potential of seven 7(11y + 29)=9(sixty 5+9y) for this reason: 77y + 203 = 585+81y subtract 77y 203 = 585 + 4y subtract 585 -382=4y divide by potential of four -ninety 5.5 = y cool, we’ve y then we replace y in equation 4) to get x so: 4) x=(sixty 5+9y)/7 x=(sixty 5 + 9*-ninety 5.5)/7 x=(sixty 5 – 859.5)/7 x=-794.7/7 x=-113.5 so we’ve x=-113.5 and y=-ninety 5.5 we now replace the two values on equation 2) so: 2) x – 5y + z = 15 -113.5 – 5*-ninety 5.5 + z = 15 -113.5 + 477.5 + z = 15 364 + z = 15 subtract 364 z = -349 so we’ve: x=-113.5, y=-ninety 5.5 and z = -349 so: (x, y, z) (-113.5, -ninety 5.5, -349)

ok you need 2 equations.

A (adults) + S (students) = 400

3A + 2S = 1050

Then you solve for A and for S

i just did it through trial and error there where 150 sudents and 250 adults sorry i cant give you a better answer

well the equation would look like:

2s + 3a = 1050

thats what i believe. if thats correct, than you have to guess and check.