Let's solve the two systems of linear equations step-by-step.

(a)

The system of equations is:

1. $3x + 4y - 23 = 0$
2. $2x - 3y - 4 = 0$

Step 1: Solve Equation (1) for $x$

Rearrange equation (1) to isolate $x$:

$3x + 4y - 23 = 0 \implies 3x = 23 - 4y \implies x = \frac{23 - 4y}{3}$

Step 2: Substitute $x$ into Equation (2)

Substitute $x = \frac{23 - 4y}{3}$ into equation (2):

$2x - 3y - 4 = 0$ $2\left( \frac{23 - 4y}{3} \right) - 3y - 4 = 0$ Multiply both terms inside the parentheses by 2: $\frac{2(23 - 4y)}{3} - 3y - 4 = 0 \implies \frac{46 - 8y}{3} - 3y - 4 = 0$ Multiply through by 3 to eliminate the denominator: $46 - 8y - 9y - 12 = 0 \implies 46 - 12 - 17y = 0 \implies 34 - 17y = 0$

Step 3: Solve for $y$

$34 - 17y = 0 \implies 17y = 34 \implies y = \frac{34}{17} = 2$

Step 4: Substitute $y = 2$ into Equation (1)

Substitute $y = 2$ back into equation (1) to find $x$:

$3x + 4(2) - 23 = 0 \implies 3x + 8 - 23 = 0 \implies 3x - 15 = 0 \implies 3x = 15 \implies x = \frac{15}{3} = 5$

Solution for (a):

The solution is $x = 5$ and $y = 2$.

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(b)

The system of equations is:

1. $5x - 4y = 7$
2. $2x + y = 8$

Step 1: Solve Equation (2) for $y$

Rearrange equation (2) to isolate $y$:

$2x + y = 8 \implies y = 8 - 2x$

Step 2: Substitute $y = 8 - 2x$ into Equation (1)

Substitute $y = 8 - 2x$ into equation (1):

$5x - 4y = 7$ $5x - 4(8 - 2x) = 7$ Distribute the $-4$: $5x - 32 + 8x = 7 \implies 13x - 32 = 7$

Step 3: Solve for $x$

$13x - 32 = 7 \implies 13x = 39 \implies x = \frac{39}{13} = 3$

Step 4: Substitute $x = 3$ into Equation (2)

Substitute $x = 3$ back into equation (2) to find $y$:

$2(3) + y = 8 \implies 6 + y = 8 \implies y = 8 - 6 = 2$

Solution for (b):

The solution is $x = 3$ and $y = 2$.

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Final Solutions:

• For (a): $x = 5$ and $y = 2$
• For (b): $x = 3$ and $y = 2$

Let me know if you need further clarification or details!

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Further questions to explore:

1. What are the graphical representations of these systems of equations?
2. How can substitution be used in other systems of equations with different coefficients?
3. How does the method of elimination compare to substitution in solving systems of equations?
4. Can these systems of equations be solved using matrices and determinants? How?
5. What would happen if these were quadratic equations instead of linear ones?

Tip: When solving systems of equations, always check if the solution satisfies both original equations to confirm your results.