How do you substitute to determine if the ordered pair (3, 2) is a solution of the system of equations y=-x+5 and x-2y=-4?

3 Answers

Answer:

#(3, 2)# isn't a solution of the system of equations.

Explanation:

You substitute the new thing for the old thing,
and you replace the old thing with or by the new thing.

Substitute 3 for x and 2 for y, and check if both equations are correct?
#y=-x+5 and x-2y=-4# & # x = 3, y = 2:#

Is #3 -2 xx2 = -4# ?
Is #-1 = -4#? No!!

Is this true #2 = -3 + 5#?
#2 = 2# , it's true

(3,2) lies on one the line but not both, and it is not the not a solution of the system of equations.

https://www.desmos.com/calculator/hw8eotboqh

Mar 26, 2018

Answer:

See Below.

Explanation:

In an ordered pair #(x,y)#; The first term is the value for the first

variable and the second term is the value for the second variable in

a system of simultaneous equations.

So, Here, We have, #(3,2)# as an ordered pair.

And, The Equations:

#y = -x + 5#..........................(i)

#x - 2y = -4#...........................(ii)

Let's substitute #x = 3# and #y = 2# in the equations eq(i) and eq(ii).

For (i):

#2 = -3 + 5# Which is correct, So The ordered pair satisfies this equation.

For (ii):

#3 - 4 = -4# Which is not possible, So, The ordered pair does not satisfy the equation.

So, The ordered pair #(3,2)# isn't a solution for this system of simultaneous equations.

Hope this helps.

Mar 26, 2018

Answer:

#(3,2)# is not the solution.

The solution is #(2,3)#.

Explanation:

#"Equation 1":# #y=-x+5#

#"Equation 2":# #x-2y=-4#

Since Equation 1 is already solved for #y#, substitute #color(red)(-x+5)# for #y# in Equation 2 and solve for #x#.

#x-2(color(red)(-x+5))=-4#

Expand.

#x+2x-10=-4#

Simplify.

#3x-10=-4#

Add #10# to both sides.

#3x=-4+10#

Simplify.

#3x=6#

Divide both sides by #3#.

#x=6/3#

#color(blue)(x=2#

Now substitute #color(blue)(2# for #x# in Equation 1 and solve for #y#.

#y=-color(blue)(2)+5#

#color(green)(y=3#

The solution is #(2,3)#, therefore #(3,2)# is not the solution.

graph{(y+x-5)(x-2y+4)=0 [-10, 10, -5, 5]}