Algebra II: Solve linear System?

x + y = -5

MULTIPLY BY 4

4x + 4y = -20

3x - 4y = -1

ADD

7x = -21

x = -3

y = -2 ANSWER

Solve the second equation for y:

x + y = -5

y = -x - 5

Substitute that into the first equation and solve for x:

3x - 4y = -1

3x - 4(-x - 5) = -1

3x + 4x + 20 = -1

7x + 20 = -1

7x = -21

x = -3

Plug that back into the equation y = -x - 5 and solve for y:

y = -(-3) - 5

y = 3 - 5

y = -2

So, the solution is x = -3, y = -2. Hope that helps :)

3x - 4y = - 1

or , 3x = - 1 + 4y -- (i)

x + y = - 5

or , 3x + 3y = -15 [ multilplying both sides by 3 ]

or, -1 + 4y + 3y = - 15 [ substituting the value of 3x from eq. (i) ]

or , 7y = - 14

or , y = - 2

Putting the value of y as -2 in the equation x + y = -5 we get x -2 = - 5

or, x = - 5 + 2 = - 3

Therefore x = - 3 and y = -2

3x -4y = -1

x + y = -5

x = -y -5

3(-y -5) -4y = -1

-3y -15 -4y = -1

-14 = 7y

y= -2

x = -3

First, use the Distributive sources on both equations: .5x + .5y + a million = 0 .5x - .5y = 2 Now, take both equation--i visit apply the bottom one--to treatment for a variable (i visit choose upon x). .5x - .5y = 2 .5x = 2 + .5y Divide all aspects with the help of one/2. X = a million + y Now change x into the common equation: .5x + .5y + a million = 0 .5(a million + y) + .5y + a million = 0 Use the Distributive sources. .5 + .5y + .5y + a million = 0 upload like words. a million.5 + y = 0 y = -a million.5 Now option y into the common equation: .5x + .5y + a million = 0 .5x + .5(-a million.5) + a million = 0 .5x - .75 + a million = 0 upload like words. .5x + .25 = 0 .5x = -.25 Divide both side by 1/2 of. X = -.5 ok! So now you've gotten y = -a million.5 and x = -.5!

x=-3

y=-2