Verified Answer :

  • Multiply the equations to align coefficients of \(y\) as additive inverses: \(25x - 15y = -35\) and \(6x + 15y = 66\).
  • Add these equations to eliminate \(y\) and solve for \(x\), which gives \(31x = 31\), so \(x = 1\).
  • Substitute \(x = 1\) into one of the original equations, \(2x + 5y = 22\), to find \(y = 4\).
  • The solution to the system is \(\boxed{x=1, y=4}\).

Explanation

  1. Analysis of the System of Equations

We are presented with a system of two linear equations in two variables, \(x\) and \(y\):
\[ \begin{cases} 5x - 3y = -7 \quad \text{(Equation 1)} \ 2x + 5y = 22 \quad \text{(Equation 2)} \end{cases} \]
Our objective is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. We can achieve this by using the method of elimination, which involves manipulating the equations so that one of the variables is canceled out when the equations are added or subtracted.

  1. Eliminating the Variable \(y\)

To eliminate the variable \(y\), we look for a common multiple of the coefficients of \(y\), which are \(-3\) and \(5\). The least common multiple is 15. We multiply Equation 1 by 5 and Equation 2 by 3 to obtain coefficients that are additive inverses:
Multiplying Equation 1 by 5:
\[5(5x - 3y) = 5(-7) \implies 25x - 15y = -35\]
Multiplying Equation 2 by 3:
\[3(2x + 5y) = 3(22) \implies 6x + 15y = 66\]
Now we have:
\[ \begin{cases} 25x - 15y = -35 \ 6x + 15y = 66 \end{cases} \]
Adding these two equations together results in:
\[(25x + 6x) + (-15y + 15y) = -35 + 66\]
\[31x = 31\]
Dividing by 31, we find:
\[x = 1\]

  1. Solving for \(y\) and Verification

Now that we have the value of \(x = 1\), we substitute it back into one of the original equations to solve for \(y\). Using Equation 2:
\[2(1) + 5y = 22\]
\[2 + 5y = 22\]
Subtracting 2 from both sides:
\[5y = 20\]
Dividing by 5:
\[y = 4\]
We can verify this result by substituting both \(x = 1\) and \(y = 4\) into Equation 1:
\[5(1) - 3(4) = 5 - 12 = -7\]
This matches the original equation, confirming that our solution is correct.

Examples

Systems of equations are used frequently in economics and logistics. For instance, if you are a manager deciding how to combine two types of raw materials to meet a specific budget and weight requirement for a product, you would set up a system of equations. Each material has a cost and a weight contribution; by solving for the amount of each material (like \(x\) and \(y\)), you ensure your production remains cost-effective and meets engineering standards. This is exactly how the mathematical model provides a precise solution to balance constraints efficiently.