Answer :
Certainly! Let's solve the given system of linear equations step-by-step:
[tex]\[ \begin{cases} 3m + 5n = 20 \\ 4m - 6n = 30 \end{cases} \][/tex]
1. Label the equations for reference:
[tex]\[ \text{Equation 1: } 3m + 5n = 20 \][/tex]
[tex]\[ \text{Equation 2: } 4m - 6n = 30 \][/tex]
2. Choose an elimination strategy:
We will eliminate one of the variables by making the coefficients of either [tex]\(m\)[/tex] or [tex]\(n\)[/tex] equal in both equations.
To eliminate variable [tex]\(n\)[/tex], we can multiply the first equation by 6 and the second equation by 5 since the least common multiple of 5 and 6 is 30.
3. Multiply equation 1 by 6:
[tex]\[ 6(3m + 5n) = 6 \cdot 20 \][/tex]
[tex]\[ 18m + 30n = 120 \quad \text{(Equation 3)} \][/tex]
4. Multiply equation 2 by 5:
[tex]\[ 5(4m - 6n) = 5 \cdot 30 \][/tex]
[tex]\[ 20m - 30n = 150 \quad \text{(Equation 4)} \][/tex]
5. Add Equation 3 and Equation 4 to eliminate [tex]\(n\)[/tex]:
[tex]\[ (18m + 30n) + (20m - 30n) = 120 + 150 \][/tex]
[tex]\[ 18m + 20m + 30n - 30n = 270 \][/tex]
[tex]\[ 38m = 270 \][/tex]
6. Solve for [tex]\(m\)[/tex]:
[tex]\[ m = \frac{270}{38} \][/tex]
[tex]\[ m = \frac{135}{19} \][/tex]
7. Substitute [tex]\(m = \frac{135}{19}\)[/tex] back into one of the original equations to solve for [tex]\(n\)[/tex]. We'll use Equation 1:
[tex]\[ 3m + 5n = 20 \][/tex]
[tex]\[ 3\left(\frac{135}{19}\right) + 5n = 20 \][/tex]
[tex]\[ \frac{405}{19} + 5n = 20 \][/tex]
8. Isolate [tex]\(5n\)[/tex]:
[tex]\[ 5n = 20 - \frac{405}{19} \][/tex]
[tex]\[ 5n = \frac{380}{19} - \frac{405}{19} \][/tex]
[tex]\[ 5n = \frac{380 - 405}{19} \][/tex]
[tex]\[ 5n = \frac{-25}{19} \][/tex]
9. Solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{-25}{5 \cdot 19} \][/tex]
[tex]\[ n = \frac{-25}{95} \][/tex]
[tex]\[ n = \frac{-5}{19} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ m = \frac{135}{19}, \quad n = \frac{-5}{19} \][/tex]
This means the values of [tex]\(m\)[/tex] and [tex]\(n\)[/tex] that satisfy both equations are:
[tex]\[ m = \frac{135}{19}, \quad n = \frac{-5}{19} \][/tex]
[tex]\[ \begin{cases} 3m + 5n = 20 \\ 4m - 6n = 30 \end{cases} \][/tex]
1. Label the equations for reference:
[tex]\[ \text{Equation 1: } 3m + 5n = 20 \][/tex]
[tex]\[ \text{Equation 2: } 4m - 6n = 30 \][/tex]
2. Choose an elimination strategy:
We will eliminate one of the variables by making the coefficients of either [tex]\(m\)[/tex] or [tex]\(n\)[/tex] equal in both equations.
To eliminate variable [tex]\(n\)[/tex], we can multiply the first equation by 6 and the second equation by 5 since the least common multiple of 5 and 6 is 30.
3. Multiply equation 1 by 6:
[tex]\[ 6(3m + 5n) = 6 \cdot 20 \][/tex]
[tex]\[ 18m + 30n = 120 \quad \text{(Equation 3)} \][/tex]
4. Multiply equation 2 by 5:
[tex]\[ 5(4m - 6n) = 5 \cdot 30 \][/tex]
[tex]\[ 20m - 30n = 150 \quad \text{(Equation 4)} \][/tex]
5. Add Equation 3 and Equation 4 to eliminate [tex]\(n\)[/tex]:
[tex]\[ (18m + 30n) + (20m - 30n) = 120 + 150 \][/tex]
[tex]\[ 18m + 20m + 30n - 30n = 270 \][/tex]
[tex]\[ 38m = 270 \][/tex]
6. Solve for [tex]\(m\)[/tex]:
[tex]\[ m = \frac{270}{38} \][/tex]
[tex]\[ m = \frac{135}{19} \][/tex]
7. Substitute [tex]\(m = \frac{135}{19}\)[/tex] back into one of the original equations to solve for [tex]\(n\)[/tex]. We'll use Equation 1:
[tex]\[ 3m + 5n = 20 \][/tex]
[tex]\[ 3\left(\frac{135}{19}\right) + 5n = 20 \][/tex]
[tex]\[ \frac{405}{19} + 5n = 20 \][/tex]
8. Isolate [tex]\(5n\)[/tex]:
[tex]\[ 5n = 20 - \frac{405}{19} \][/tex]
[tex]\[ 5n = \frac{380}{19} - \frac{405}{19} \][/tex]
[tex]\[ 5n = \frac{380 - 405}{19} \][/tex]
[tex]\[ 5n = \frac{-25}{19} \][/tex]
9. Solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{-25}{5 \cdot 19} \][/tex]
[tex]\[ n = \frac{-25}{95} \][/tex]
[tex]\[ n = \frac{-5}{19} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ m = \frac{135}{19}, \quad n = \frac{-5}{19} \][/tex]
This means the values of [tex]\(m\)[/tex] and [tex]\(n\)[/tex] that satisfy both equations are:
[tex]\[ m = \frac{135}{19}, \quad n = \frac{-5}{19} \][/tex]