Help solving a system of equations using the substitution method.?

Question

I need help with the following two questions:

1: y=2x-8
4x+3y=1

2. y=x-5
3x-4(y-2)=28-x

Thanks in advance for the help!

Answer 1

Question 1
4x + 3(2x - 8) = 1
10x - 24 = 1
10x = 25
x = 2.5
y = 5 - 8
y = - 3
Answer is x = 2.5 , y = - 3

Question 2
3x - 4(2x - 10) = 28 - x
3x - 8x + 40 = 28 - x
12 = 4x
x = 3
y = - 2

Answer 2

1) y = 2x-8 and
4x+3y+1. Since we have an expression for y (=2x-8), we SUBSTITUTE it this way:

4x + 3(2x-8) = 1, expand

4x + 6x -24 = 1, gather like terms

4x + 6x = 1+24 (note sign change)

10x = 25, divide by 10

x= 25/10 or x=2.5.

So, what is y?

use y = 2x-8, and use x=2.5 or
y = 2(2.5) -8
= 5 -8
y = -3.

Solution is the POINT (2.5, -3).

CHECK by substituting both into

4x+3y=1 or

4(2.5) + 3(-3) =? 1 (=? means is it really equal to?)
10 -9 =? 1. check.

2) same thing here, substitute y = x-5 into
3x -4(y-2) = 28-x, or
3x -4((x-5)-2) = 28-x. see what happened to the y? simplify
3x -4(x-7) = 28-x expand
3x -4x +28 = 28-x. gather like terms
3x-4x+x = 28-28. Note sign changes.
0 = 0. interesting!

Expand second equaqtion as 3x -4y+8 = 28-x, solve for y
-4y = 28-x -3x -8 or
-4y = -4x + 20. Divide by (-4)
y = x -5.

You see, they are the SAME line, hence they have an INFINITE number of intersection points!

Answer 3

1. y = 2x-8
4x + 3(2x-8) = 1
4x + 6x - 24 = 1
10x = 25
x = 5/2
y = 2(5/2)-8 = 5 - 8 = -3

2. y = x-5
3x - 4[(x-5)-2] = 28 - x
3x - 4[x - 7] = 28 - x
3x - 4x + 28 = 28 - x
-x = -x
all real x are solutions.
2nd equation simplifies to
3x - 4[(y-2] = 28 - x
3x - 4y + 8 = 28 - x
-4y = -4x + 20
y = x - 5

Answer 4

1) sub first into second
4x+3(2x-8)=1
10x-24=1
10x=25
x=2.5
y=2(2.5)-8 = -3

2)like the 1st one
3x - 4(2x-8-2)=28-x
3x - 8x+40=28-x
-5x+40=28-x
12=4x
x=3
y=-2

Answer 5

1: y=2x-8
4x+3y=1
substitute
4x+3(2x-8)=1
4x+6x-24=1
10x=25
x=5/2
y=2(5/2)-8
= -3

Similarly try and solve the other one