Systems Substitution 32B47E

1. Solve the system by substitution: Given: \( y = -x + 4 \) and \( y = 3x \). Set the two expressions for \( y \) equal: $$ -x + 4 = 3x $$ Add \( x \) to both sides: $$ \cancel{-x} + 4 + \cancel{x} = 3x + x \implies 4 = 4x $$ Divide both sides by 4: $$ \frac{4}{\cancel{4}} = \frac{4x}{\cancel{4}} \implies 1 = x $$ Substitute \( x = 1 \) into \( y = 3x \): $$ y = 3(1) = 3 $$ Solution: \( (1, 3) \). 2. Given: \( y = 2x - 10 \) and \( 2y = x - 8 \). Substitute \( y \) into second equation: $$ 2(2x - 10) = x - 8 $$ Simplify: $$ 4x - 20 = x - 8 $$ Subtract \( x \) from both sides: $$ 4x - 20 - x = x - 8 - x \implies 3x - 20 = -8 $$ Add 20 to both sides: $$ 3x - 20 + 20 = -8 + 20 \implies 3x = 12 $$ Divide both sides by 3: $$ \frac{3x}{\cancel{3}} = \frac{12}{\cancel{3}} \implies x = 4 $$ Substitute \( x = 4 \) into \( y = 2x - 10 \): $$ y = 2(4) - 10 = 8 - 10 = -2 $$ Solution: \( (4, -2) \). 3. Given: \( x - 2y = 12 \) and \( y = 3x + 14 \). Substitute \( y \) into first equation: $$ x - 2(3x + 14) = 12 $$ Simplify: $$ x - 6x - 28 = 12 $$ Combine like terms: $$ -5x - 28 = 12 $$ Add 28 to both sides: $$ -5x = 40 $$ Divide both sides by -5: $$ \frac{-5x}{\cancel{-5}} = \frac{40}{\cancel{-5}} \implies x = -8 $$ Substitute \( x = -8 \) into \( y = 3x + 14 \): $$ y = 3(-8) + 14 = -24 + 14 = -10 $$ Solution: \( (-8, -10) \). 4. Given: \( x = 2y - 6 \) and \( y = 3x - 7 \). Substitute \( x \) into second equation: $$ y = 3(2y - 6) - 7 $$ Simplify: $$ y = 6y - 18 - 7 $$ $$ y = 6y - 25 $$ Subtract \( 6y \) from both sides: $$ y - 6y = -25 \implies -5y = -25 $$ Divide both sides by -5: $$ \frac{-5y}{\cancel{-5}} = \frac{-25}{\cancel{-5}} \implies y = 5 $$ Substitute \( y = 5 \) into \( x = 2y - 6 \): $$ x = 2(5) - 6 = 10 - 6 = 4 $$ Solution: \( (4, 5) \). 5. Given: \( 6x - 4y = 18 \) and \( -x - 6y = 7 \). Solve second for \( x \): $$ -x - 6y = 7 \implies -x = 7 + 6y \implies x = -7 - 6y $$ Substitute into first equation: $$ 6(-7 - 6y) - 4y = 18 $$ Simplify: $$ -42 - 36y - 4y = 18 $$ Combine like terms: $$ -42 - 40y = 18 $$ Add 42 to both sides: $$ -40y = 60 $$ Divide both sides by -40: $$ \frac{-40y}{\cancel{-40}} = \frac{60}{\cancel{-40}} \implies y = -\frac{3}{2} $$ Substitute \( y = -\frac{3}{2} \) into \( x = -7 - 6y \): $$ x = -7 - 6\left(-\frac{3}{2}\right) = -7 + 9 = 2 $$ Solution: \( (2, -\frac{3}{2}) \). 6. Given: \( 9x - 3y = 9 \) and \( 3x - y = 3 \). Solve second for \( y \): $$ 3x - y = 3 \implies y = 3x - 3 $$ Substitute into first equation: $$ 9x - 3(3x - 3) = 9 $$ Simplify: $$ 9x - 9x + 9 = 9 $$ $$ 9 = 9 $$ This is true for all \( x \), so infinite solutions along \( y = 3x - 3 \). 7. Given: \( y = 3x + 8 \) and \( 2y = 6x + 16 \). Substitute \( y \) into second equation: $$ 2(3x + 8) = 6x + 16 $$ Simplify: $$ 6x + 16 = 6x + 16 $$ True for all \( x \), infinite solutions along \( y = 3x + 8 \). 8. Given: \( y = 4x + 5 \) and \( 12x - 3y = 9 \). Substitute \( y \) into second equation: $$ 12x - 3(4x + 5) = 9 $$ Simplify: $$ 12x - 12x - 15 = 9 $$ $$ -15 = 9 $$ False, no solution. 9. Given: \( 7y = -2x + 5 \) and \( 3x + 10y = 6 \). Rewrite first as: $$ y = \frac{-2x + 5}{7} $$ Substitute into second: $$ 3x + 10\left(\frac{-2x + 5}{7}\right) = 6 $$ Multiply both sides by 7: $$ 7(3x) + 10(-2x + 5) = 42 $$ $$ 21x - 20x + 50 = 42 $$ $$ x + 50 = 42 $$ $$ x = -8 $$ Substitute \( x = -8 \) into \( y = \frac{-2x + 5}{7} \): $$ y = \frac{-2(-8) + 5}{7} = \frac{16 + 5}{7} = \frac{21}{7} = 3 $$ Solution: \( (-8, 3) \). 10. Solve by substitution: Given: \( x + y = 6 \) and \( 5x - y = 3 \). From first: \( y = 6 - x \). Substitute into second: $$ 5x - (6 - x) = 3 $$ Simplify: $$ 5x - 6 + x = 3 $$ $$ 6x - 6 = 3 $$ Add 6: $$ 6x = 9 $$ Divide by 6: $$ x = \frac{9}{6} = \frac{3}{2} $$ Substitute into \( y = 6 - x \): $$ y = 6 - \frac{3}{2} = \frac{12}{2} - \frac{3}{2} = \frac{9}{2} $$ Solution: \( \left(\frac{3}{2}, \frac{9}{2}\right) \). Graphing method involves plotting both lines and finding their intersection visually. Substitution is more accurate as it gives exact values. 11. Let \( a \) = number of adult tickets, \( c \) = number of children tickets. Given: $$ a + c = 400 $$ $$ 8a + 5c = 2750 $$ Solve first for \( c \): $$ c = 400 - a $$ Substitute into second: $$ 8a + 5(400 - a) = 2750 $$ Simplify: $$ 8a + 2000 - 5a = 2750 $$ $$ 3a + 2000 = 2750 $$ Subtract 2000: $$ 3a = 750 $$ Divide by 3: $$ a = 250 $$ Substitute into \( c = 400 - a \): $$ c = 400 - 250 = 150 $$ Solution: 250 adult tickets and 150 children tickets sold.