How can you solve the system of equations represented by 7x + 2y = 4 and 5y + 3x = 10 algebraically?

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To solve the system of equations given by 7x + 2y = 4 and 5y + 3x = 10, we can use the method of substitution or elimination. In this solution, we will employ the elimination method for clarity.

First, let’s rewrite the equations clearly:

  1. Equation 1: 7x + 2y = 4
  2. Equation 2: 3x + 5y = 10

To eliminate one variable, we can make the coefficients of y the same in both equations. We will multiply Equation 1 by 5 and Equation 2 by 2:

  • 5(7x + 2y) = 5(4) ⇒ 35x + 10y = 20 (New Equation 1)
  • 2(3x + 5y) = 2(10) ⇒ 6x + 10y = 20 (New Equation 2)

Now we have:

  1. 35x + 10y = 20 (New Equation 1)
  2. 6x + 10y = 20 (New Equation 2)

Next, we can subtract New Equation 2 from New Equation 1 to eliminate y:

(35x + 10y) - (6x + 10y) = 20 - 20

This simplifies to:

29x = 0

Now, we can solve for x: 

x = 0

With the value of x, we can substitute back into either original equation to find y. Using Equation 1:

7(0) + 2y = 4

This simplifies to:

2y = 4
y = 2

Thus, the solution to the system of equations is:

x = 0 and y = 2.

If we want to verify our solution, we can substitute x and y back into the second equation:

3(0) + 5(2) = 10

Which simplifies to:

10 = 10

This confirms that our solution is indeed correct.

In summary, the system of equations 7x + 2y = 4 and 5y + 3x = 10 has the unique solution:

x = 0, y = 2.

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