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Sagot :
To solve the system of equations and find the value of [tex]\(x - y\)[/tex] for both given sets of equations, we'll proceed with solving each system step-by-step.
### Part (i):
Given equations:
[tex]\[ \begin{cases} 5x + 4y = 14 \quad \text{(1)} \\ 4x + 5y = 13 \quad \text{(2)} \end{cases} \][/tex]
First, let's solve the system:
1. Multiply equation (1) by 5 and equation (2) by 4 to make the coefficients of [tex]\( y \)[/tex] the same:
[tex]\[ \begin{cases} 25x + 20y = 70 \quad \text{(3)} \\ 16x + 20y = 52 \quad \text{(4)} \end{cases} \][/tex]
2. Subtract equation (4) from equation (3):
[tex]\[ (25x + 20y) - (16x + 20y) = 70 - 52 \\ 25x - 16x = 18 \\ 9x = 18 \\ x = 2 \][/tex]
3. Substitute [tex]\( x = 2 \)[/tex] back into equation (1) to find [tex]\( y \)[/tex]:
[tex]\[ 5(2) + 4y = 14 \\ 10 + 4y = 14 \\ 4y = 4 \\ y = 1 \][/tex]
4. Now, calculate [tex]\( x - y \)[/tex]:
[tex]\[ x - y = 2 - 1 = 1 \][/tex]
### Part (ii):
Given equations:
[tex]\[ \begin{cases} 2x - 5y = 5 \quad \text{(1)} \\ 5x - 8y = 14 \quad \text{(2)} \end{cases} \][/tex]
First, let's solve the system:
1. Multiply equation (1) by 5 and equation (2) by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[ \begin{cases} 10x - 25y = 25 \quad \text{(3)} \\ 10x - 16y = 28 \quad \text{(4)} \end{cases} \][/tex]
2. Subtract equation (4) from equation (3):
[tex]\[ (10x - 25y) - (10x - 16y) = 25 - 28 \\ -25y + 16y = -3 \\ -9y = -3 \\ y = \frac{1}{3} \][/tex]
3. Substitute [tex]\( y = \frac{1}{3} \)[/tex] back into equation (1) to find [tex]\( x \)[/tex]:
[tex]\[ 2x - 5\left(\frac{1}{3}\right) = 5 \\ 2x - \frac{5}{3} = 5 \\ 2x = 5 + \frac{5}{3} \\ 2x = \frac{15}{3} + \frac{5}{3} \\ 2x = \frac{20}{3} \\ x = \frac{10}{3} \][/tex]
4. Now, calculate [tex]\( x - y \)[/tex]:
[tex]\[ x - y = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3 \][/tex]
### Conclusion
After solving both systems, we find:
[tex]\[ \begin{aligned} (i) &\quad x - y = 1 \\ (ii) &\quad x - y = 3 \end{aligned} \][/tex]
Thus, the values of [tex]\( x - y \)[/tex] are [tex]\( 1 \)[/tex] and [tex]\( 3 \)[/tex] respectively for the two systems of equations.
### Part (i):
Given equations:
[tex]\[ \begin{cases} 5x + 4y = 14 \quad \text{(1)} \\ 4x + 5y = 13 \quad \text{(2)} \end{cases} \][/tex]
First, let's solve the system:
1. Multiply equation (1) by 5 and equation (2) by 4 to make the coefficients of [tex]\( y \)[/tex] the same:
[tex]\[ \begin{cases} 25x + 20y = 70 \quad \text{(3)} \\ 16x + 20y = 52 \quad \text{(4)} \end{cases} \][/tex]
2. Subtract equation (4) from equation (3):
[tex]\[ (25x + 20y) - (16x + 20y) = 70 - 52 \\ 25x - 16x = 18 \\ 9x = 18 \\ x = 2 \][/tex]
3. Substitute [tex]\( x = 2 \)[/tex] back into equation (1) to find [tex]\( y \)[/tex]:
[tex]\[ 5(2) + 4y = 14 \\ 10 + 4y = 14 \\ 4y = 4 \\ y = 1 \][/tex]
4. Now, calculate [tex]\( x - y \)[/tex]:
[tex]\[ x - y = 2 - 1 = 1 \][/tex]
### Part (ii):
Given equations:
[tex]\[ \begin{cases} 2x - 5y = 5 \quad \text{(1)} \\ 5x - 8y = 14 \quad \text{(2)} \end{cases} \][/tex]
First, let's solve the system:
1. Multiply equation (1) by 5 and equation (2) by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[ \begin{cases} 10x - 25y = 25 \quad \text{(3)} \\ 10x - 16y = 28 \quad \text{(4)} \end{cases} \][/tex]
2. Subtract equation (4) from equation (3):
[tex]\[ (10x - 25y) - (10x - 16y) = 25 - 28 \\ -25y + 16y = -3 \\ -9y = -3 \\ y = \frac{1}{3} \][/tex]
3. Substitute [tex]\( y = \frac{1}{3} \)[/tex] back into equation (1) to find [tex]\( x \)[/tex]:
[tex]\[ 2x - 5\left(\frac{1}{3}\right) = 5 \\ 2x - \frac{5}{3} = 5 \\ 2x = 5 + \frac{5}{3} \\ 2x = \frac{15}{3} + \frac{5}{3} \\ 2x = \frac{20}{3} \\ x = \frac{10}{3} \][/tex]
4. Now, calculate [tex]\( x - y \)[/tex]:
[tex]\[ x - y = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3 \][/tex]
### Conclusion
After solving both systems, we find:
[tex]\[ \begin{aligned} (i) &\quad x - y = 1 \\ (ii) &\quad x - y = 3 \end{aligned} \][/tex]
Thus, the values of [tex]\( x - y \)[/tex] are [tex]\( 1 \)[/tex] and [tex]\( 3 \)[/tex] respectively for the two systems of equations.
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