solve 5x+y=8 and 3x-4y=14?

Question

i cant figure it out! :/
I need to solve by graphing,
than substitution,
than elimination.
3 different answers.

helpppp please!

Answer 1

5x + y = 8

Subtract 5x to the other side so you have y = -5x + 8.

3x - 4y = 14 subtract 3x over:

=> -4y = -3x + 14 divide by -4 to get:

=> y = (3x - 14) / 4

Now you have two equations with respect to x

The equations fit the form y = mx + b, where b is your starting point on the y-axis and m is the rate at which the line goes up (rise/run). Put a dot at the value of b. In the case of the second equation, you will be going up three squares, then four to the right. Place a dot there. Your y-intercept (b value) is going to be 7/2, so just sort guesstimate halfway between three and four.

You're going to do the same for the first equation.

The point at which they meet is the point of intersection (which is what I am assuming you are looking for?).

Using the substitution method, all you are doing is plugging one equation into the other and solving for the variable. All you have to do then is plug the number back into either one of the two equations to get the x and y coordinates for the point.

If that doesn't do it for you, I apologize. I'm going to try and get my yahoo instant messenger working, so if you need more help, drop me a line within the next couple of minutes...

Answer 2

To solve by graphing, plot each line on graph paper. The point where the lines intersect is the solution.

5x+y=8 and 3x-4y=14

Substitution:
Easiest is the 1st equation:
y = 8 - 5x
Then in the 2nd equation,
3x - 4(8-5x) = 14
Solve for x, then substitute for x in one of the original equations and solve for y.

Elimination:

5x + y = 8
..3x - 4y = 14, Multiply the 1st equation by 4:
20x + 4y = 32, Add:
23x.........= 46
solve for x, substitute back and solve for y

Answer 3

if you have a graphing calc type in the equations and the point at which they intersect is you answer.

Substitution: solve for y in 5x+y=8 so Y=(8-5x) then plug that into the other equation and solve for y so y should be -2 then plug that in for y and get x=2

Elimination: Get the ys to be opposite of each other.
5x+y=8 and 3x-4y=14

leave the second one alone and multiply the first by 4 so that
20x+4y=32

20x+4y=32
3x -4y=14

add together to get 23x=46 so x=2 then plug that in to get -2

Answer 4

Graph Method For the first eqn. y=8- 5x .Therefore, for x= 2,4 and 6 ],the value of y would respectively be -2, -12 and -22 . For the second equation 3x- 4y=14 => -4y=14-3x =>4y=3x-14 =>y=(3x-14)/4 . Therefore for values of x=2,6 and 10 value of y would be -2. 1 and 4.Now draw two straight lines with (2.-2), (4, -12),(6, -22) AND (2, -2),(6,1),(10,4).From their point of intersection,you may find the value of x=2 and y= -2. NOW LET US DO IT BY SUBSTUTITUTION METHOD. From the first equation we get y= 8-5x.now putting this value of y in the second equation we get. 3x-4(8-5x)=14 => 3x-32+20x=14 => 23x=14+32=46 =>x=46/23=2. Putting the value of x in the eqn.y=8-5x,we get the value of y=-2 Thereforex=2 and y=- 2 ans NOW LET US SOLVE THE SAME PROBLEM BY ELLIMINATION METHOD Multiplying the first eqn. by 3 and the second equation by 5 we get 15x+3y=24 and 15x-20y=70 Subtracting ,we get 23y=- 46 => y=- 46/23=- 2.putting the value of y in eqn. we can find the value of x=2 .So are you happy now that by all the processes the answer is same.And it should be.

Answer 5

5x+y = 8 => y = 8 - 5x

3x - 4(8-5x) = 14

3x - 32 - 20x = 14

23x - 32 = 14

23x = 46

x = 2

5(2) + y = 8

y = -2

(2, -2)

10 + y = 8

y = -2

x = 2

Answer 6

x= 2
y= -2

For substitution, isolate the y to get "y= -5x + 8" then plug the y value in for the other eqution. Once you find out what x is (2) plug that vlaue into the first equation, and then you learn the y= -2

Answer 7

5x+y=8
3x-4y=14

20x +4y=32
3x-4y=14

23x=46
x=2, y=-2