Question

A system of equations 2x+4y=10

2x-4y=10

select the answer that has the same solution

a
2x+6y=14

2x-6y=14

b
3x-5y=15

3x+5y=15

c
8x-10y=12

8x+10y=12

Answer 1

To determine which set of equations has the same solution as the given system:

1. **Original System of Equations:**
\[
2x + 4y = 10 \quad (1)
\]
\[
2x - 4y = 10 \quad (2)
\]

First, we can simplify and find the solution for the original equations.

From equation (1):
\[
2x + 4y = 10 \implies x + 2y = 5 \quad (3)
\]

From equation (2):
\[
2x - 4y = 10 \implies x - 2y = 5 \quad (4)
\]

Now we have:
\[
x + 2y = 5 \quad (3)
\]
\[
x - 2y = 5 \quad (4)
\]

Next, we can solve equations (3) and (4) simultaneously to find \(x\) and \(y\).

Subtracting equation (4) from equation (3):
\[
(x + 2y) - (x - 2y) = 5 - 5
\]
\[
4y = 0 \implies y = 0
\]

Substituting \(y = 0\) into equation (3):
\[
x + 2(0) = 5 \implies x = 5
\]

Thus, the solution to the original system of equations is:
\[
(x, y) = (5, 0)
\]

2. **Testing the Other Equations:**

**Option (a):**
\[
2x + 6y = 14 \quad (5)
\]
\[
2x - 6y = 14 \quad (6)
\]

From (5):
\[
2x + 6y = 14 \quad \text{dividing by 2:} \quad x + 3y = 7 \quad (7)
\]

From (6):
\[
2x - 6y = 14 \quad \text{dividing by 2:} \quad x - 3y = 7 \quad (8)
\]

Subtracting (8) from (7):
\[
(x + 3y) - (x - 3y) = 7 - 7
\]
\[
6y = 0 \implies y = 0
\]

Substituting \(y = 0\) into (7):
\[
x + 3(0) = 7 \implies x = 7
\]

Solution is \((7, 0)\), which is different.

---

**Option (b):**
\[
3x - 5y = 15 \quad (9)
\]
\[
3x + 5y = 15 \quad (10)
\]

From (9) plus (10):
\[
(3x - 5y) + (3x + 5y) = 15 + 15
\]
\[
6x = 30 \implies x = 5
\]

Substituting \(x = 5\) into (9):
\[
3(5) - 5y = 15 \implies 15 - 5y = 15 \implies 5y = 0 \implies y = 0
\]

Solution is \((5, 0)\), which matches.

---

**Option (c):**
\[
8x - 10y = 12 \quad (11)
\]
\[
8x + 10y = 12 \quad (12)
\]

From (11):
\[
8x - 10y = 12 \quad (dividing by 2) \quad 4x - 5y = 6 \quad (13)
\]

From (12):
\[
8x + 10y = 12 \quad (dividing by 2) \quad 4x + 5y = 6 \quad (14)
\]

Subtracting (14) from (13):
\[
(4x - 5y) - (4x + 5y) = 6 - 6
\]
\[
-10y = 0 \implies y = 0
\]

Substituting \(y = 0\) into (13):
\[
4x - 5(0) = 6 \implies 4x = 6 \implies x = \frac{3}{2}
\]

Solution is \((\frac{3}{2}, 0)\), which is different.

---

**Conclusion:**

The only option that has the same solution \((5, 0)\) as the original system is:

**Option b:**
\[
3x - 5y = 15
\]
\[
3x + 5y = 15
\]