Systems of Linear Equations

1. The first equation already gives $y$ in terms of $x$. Substitute $y = 2x - 1$ into the second equation:

$3x + (2x - 1) = 14$ $5x - 1 = 14$ $5x = 15$ $x = 3$

Back-substitute into $y = 2x - 1$:

$y = 2(3) - 1 = 5$

Solution: $(3, 5)$

Check: $3(3) + 5 = 9 + 5 = 14$ ✓

2. The $y$-coefficients are $+3$ and $-3$, so they cancel when added:

$(2x + 3y) + (4x - 3y) = 12 + 6$ $6x = 18$ $x = 3$

Substitute $x = 3$ into the first equation:

$2(3) + 3y = 12 \implies 6 + 3y = 12 \implies 3y = 6 \implies y = 2$

Solution: $(3, 2)$

Check: $4(3) - 3(2) = 12 - 6 = 6$ ✓

3. (a) Set the expressions equal (both equal $y$):

$x + 2 = -x + 6$ $2x = 4$ $x = 2$

$y = 2 + 2 = 4$

Intersection: $(2, 4)$

(b) The intersection point $(2, 4)$ is the unique solution to the system — the only $(x, y)$ pair that satisfies both equations simultaneously. Graphically, it is where the two lines cross.

4. Rewrite the second equation by multiplying by $2$:

$2(3x - y) = 2(5) \implies 6x - 2y = 10$

This is identical to the first equation. Both equations represent the same line, so the system has infinitely many solutions (every point on the line $3x - y = 5$ is a solution).

5. To eliminate $y$, multiply the first equation by $2$:

$2(3x + 2y) = 2(16) \implies 6x + 4y = 32$

Add to the second equation:

$(6x + 4y) + (5x - 4y) = 32 + 4$ $11x = 36$ $x = \frac{36}{11}$

Hmm — let's check for a cleaner approach using the original equations directly:

Multiply equation 1 by $2$: $6x + 4y = 32$

Add to equation 2: $(6x + 4y) + (5x - 4y) = 32 + 4 \Rightarrow 11x = 36 \Rightarrow x = \frac{36}{11}$

Substitute back into equation 1:

$3\!\left(\frac{36}{11}\right) + 2y = 16 \implies \frac{108}{11} + 2y = 16 \implies 2y = 16 - \frac{108}{11} = \frac{176-108}{11} = \frac{68}{11} \implies y = \frac{34}{11}$

Solution: $\left(\dfrac{36}{11},\ \dfrac{34}{11}\right)$

Check (equation 2): $5\!\left(\frac{36}{11}\right) - 4\!\left(\frac{34}{11}\right) = \frac{180 - 136}{11} = \frac{44}{11} = 4$ ✓

6. Let $p$ = number of pencils, $n$ = number of pens.

Equation 1 (total items): $p + n = 12$

Equation 2 (total cost): $0.25p + 0.75n = 6.00$

Multiply equation 2 by $4$ to clear decimals:

$p + 3n = 24$

Subtract equation 1 from this:

$(p + 3n) - (p + n) = 24 - 12$ $2n = 12$ $n = 6$

From equation 1: $p = 12 - 6 = 6$

The student bought 6 pencils and 6 pens.

Check: 6(0.25) + 6(0.75) = 1.50 + 4.50 = \6.00$ ✓