Solve the system of equations: 3r + 4s = 0 and 2r + 5s = 23

To solve the system of equations, we have:

• Equation 1: 3r + 4s = 0
• Equation 2: 2r + 5s = 23

We can use the substitution or elimination method to find the values of r and s. Here, we’ll use the elimination method.

First, let’s solve for one variable in terms of the other using Equation 1:

3r + 4s = 0  00  00  
=> 4s = -3r  
=> s = -3r / 4

Now, substitute the expression for s into Equation 2:

2r + 5(-3r / 4) = 23  
=> 2r - 15r / 4 = 23

To eliminate the fraction, multiply the entire equation by 4:

4(2r) - 15r = 4(23)  
=> 8r - 15r = 92

This simplifies to:

-7r = 92

Now, solving for r:

r = 92 / -7  
=> r = -13.14 (approximately)

Next, substitute r back into the equation for s:

s = -3(-13.14) / 4  
=> s = 39.42 / 4  
=> s = 9.85 (approximately)

Thus, the solution to the system of equations is:

• r ≈ -13.14
• s ≈ 9.85

These values satisfy both equations, completing the solution.