Solve the system.Enter the solution coordinates.?

Question

2x + y = 4
x + y = 3

Answer 1

x+y=3 => y=3-x
=> 2x+(3-x)=4 => x+3=4 => x=1
y=3-x & x=1 => y=2

=> x=1 & y=2

Answer 2

x = 1 and y=2

Answer 3

cchappa is correct. You should understand what it means rather than asking for solutions.

In any system of equations, what is required is to find values for the unknowns (that are the SAME in each equation) such that all the equations are TRUE.

In the above case, you have the simplest form: simultaneous equations. Well, I don't agree with the approach used by your math teachers (intersection, coincidence or parallelism of lines) because it shows a clear lack of understanding. So how does one proceed? Can you see that if you had one equation in one unknown, it would be easy to solve? This is the first step. Secondly, each equation affects the solution so each equation introduces its influence by replacing variables in one of the other equations. Thirdly, the process is repeated until we are able to solve for all the unknowns. Let's see how this works for the above system:

First, choose a variable (say y):
y = -2x + 4 [1]
y = -x + 3 [2]

If we want the same value for y in the above equations, can you not see that this is only possible if the RHS (right hand sides) of [1] and [2] are equal? So you set these equal and obtain: -2x + 4 = -x + 3 which is an equation in one unknown and thus easy to solve. Solution is: x = 1. Now you can find y by replacing x in any of the above equations. So y=2.

You could have started out with x rather than y:

x = (4 - y)/2 [3]
x = 3 - y [4]

Equating the RHS, we have: 2 -y/2 = 3 - y
and solving this for y:

y/2 = 1
So y=2.

Now in systems with more equations, the process becomes more difficult because there are more variables but it turns out that you are still finding solutions in terms of the coefficients and constant terms in each equation. Let's suppose your system of equations is:

ay - bx = c [5]
dy - ex = f [6]

y = (bx + c)/a
y = (ex + f)/d

(bx)/a +c/a = (ex)/d + f/d
=> (bd-ea)x/(ad) = (af-cd)/(ad)
=> (bd-ea)x = (af-cd)
=> x = (af-cd)/(bd-ea)

So a=1, b=-2, c=4, d=1, e=-1 and f=3

x = (1*3 - 4*1)/(-2*1-(-1*1))
x = (-1)/(-1) = 1

Get the idea? Now try this by finding a formula for y!

Answer 4

x = 1, y=2

Answer 5

If you can't solve a system as simple as this, you are doomed to repeat the math class you are currently taking. You'd do much better off asking HOW instead of just demanding solutions.

Answer 6

subtract LHS & RHS of eqn2 from eqn1

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

put in eqn 2
1 + y =3
y= 3 - 1 = 2

x,y= 1,2

hope it helps

Answer 7

x-2y=-7 x=-7+2y 5x-4y=-5 5(-7+2y)-4y=-5 -35+10y-4y=-5 6y=-5+35 6y=30 y=5 x=-7+2(5) x=3 x = 3, y = 5 in case you're youthful sufficient to no longer comprehend this, you in all likelihood are no longer accredited to apply yahoo! solutions.

Answer 8

x=1
y=2

Answer 9

2x + y = 4
x + y = 3

Mulitply bottom by -1

2x - y = 4
-x - y = -3

x = 1

x + y = 3
1 + y = 3
y = 2

ANS : (1,2)

Answer 10

2x+y=4
x+y=3
subtracting x=1
substituting y=2
solution set {1,2}