I need more help here on Solving the system by addition?

Question

2x - 4y =10 and 4x + 5y = 20
4x - 3y = 0, y = x + 1
-4x + 2y = 8, y = 2x + 3

Answer 1

For the first set:
2x=10+4y
x=5+2y plug this in to the second equation
4(5+2y) + 5y = 20
10+8y+5y = 20
13y = 10
y = 10/13
You can figure out the rest using this same concept.

Answer 2

Answer 1:
2x - 4y = 10.....(1)
4x + 5y = 20.....(2)

Multiply (1) by -2
-2(2x - 4y) = -20
- 4x + 8y = -20.....(3)

Add (3) and (2)

4x - 4x + 5y + 8y = 20-20
13y = 0
y = 0.....(4)

Substitute (4) in (1)

2x - 4y = 10
2x = 10
x = 5

x = 5, y = 0 is the solution


Answer 2:
4x - 3y = 0.....(1)
y = x + 1.....(2)

Consider (2)
y = x + 1
x - y + 1 = 0
-4(x - y + 1) = -4*0 (Multiply both sides by -4)
-4x + 4y - 4 = 0
4y - 4x = 4.....(3)

Add (3) and (1)

4x - 4x + 4y - 3y = 0 + 4
y = 4.....(4)

Substitute (4) in (2)

y = x + 1
4 = x + 1
x = 3

x = 3, y = 4 is the solution


Answer 3:
-4x + 2y = 8.....(1)
y = 2x + 3.....(2)

Consider (2)
y = 2x + 3
2x - y = -3.....(3)

WAIT! I noticed something. This system has no real solution. Here's how I found that out:

There is an extremely simple way of checking how many solutions a system of linear equations in two variables has by using the coefficients of the variables and the constants. Here is the method

There are two equations to solve. They are:

ax + by = c.....(1)
px + qy = r.....(2)

The varibles are x and y.

If a/p = b/q but does not equal to c/r, the system has no solution.

If a/p does not equal b/q, the system is satisfied only for one value of x and y (has a unique solution).

If a/p = b/q = c/r, the system has infinite solutions. That is, the graph of both equations is the same line.

In this problem, -4/2 = 2/-1 but they do not equal 8/-3. So the system has no real solution.

Answer 3

2x - 4y = 10 multiplying both sides of the equation by 5
10x - 20y = 50--------(1)

4x + 5y = 20 multiply both sides by 4
16x + 20 y = 80-------(2)

(1) + (2) =>
10x + 16x - 20y + 20y = 50 + 80
26x = 130
x = 130/26 = 5

replacing x in the first original eq. we get:
2 * 5 - 4y = 10
10 - 4y = 10
y = 0


4x - 3y = 0 -----------(1)
-x + y = 1 multiply each side by 3
-3x + 3y = 3 ----------(2)

(1) + (2) =>
x = 3
replacing in (1)
4 *3 - 3y = 0
3y = 12
y = 4


-4x + 2y = 8------------(1)
-2x + y = 3 multiply by (-2)
4x - 2y = -6 --------------(2)

(1) + (2) =>

0 = 2
impossible no solution

Answer 4

Question 1
10x - 20y = 50
16x + 20y = 80----ADD
26x = 130
x = 5
80 + 20y = 80
y = 0
Solution is x = 5, y = 0
Question 2
4x - 3y = 0
- 4x + 4y = 4----ADD
y = 4
4 = x + 1
x = 3
Solution is x = 3 , y = 4
Question 3
- 2x + y = 3
- 4x + 2y = 6
But - 4x + 2y = 8 is given---error in question?

Answer 5

1.
2x-4y=10
4x+5y=20

Multiply the first equation by -2:
-4x+8y=-20
4x+5y=20
=========
13y=0
y=0

4x+5(0)=20
4x=20
x=5

The solution set is (5,0)

2.
4x-3y=0
y=x+1

Subtract x from both sides on the second equation:
4x-3y=0
-x+y=1

Multiply the second equation by 3:
4x-3y=0
-3x+3y=3
=======
x=3

4(3)-3y=0
12-3y=0
-3y=-12
y=4

The solution set is (3,4)

3.
-4x+2y=8
y=2x+3

Subtract 2x from both sides:
-4x+2y=8
-2x+y=3

Multiply the second equation by -2:
-4x+2y=8
4x-2y=-6
=======
0=2

No solution!

I hope this helps!

Answer 6

y=0 x=5 multiply 2 times first equation then subtract second equation and you get y=o then plug it in equation and solve for x=5 check values in other equation.