Question

solve each system using elimination.

1. $2x + 3y = 9$

$x + 5y = 8$

1. $3x + y = 5$

$2x - 2y = -2$

1. $6x + 4y = 42$

$-3x + 3y = -6$

1. $3x + 2y = 17$

$2x + 5y = 26$

1. $6x - 3y = 15$

$7x + 4y = 10$

1. $5x - 9y = -43$

$3x + 8y = 68$
tell whether the system has one solution, infinitely many solutions, or no solution.

1. $9x + 8y = 15$

$9x + 8y = 30$

1. $3x + 4y = 24$

$6x + 8y = 24$

1. $5x - 3y = 10$

$10x + 6y = 20$

1. $2x - 5y = 17$

$6x - 15y = 51$

1. $4x - 7y = 15$

$-8x + 14y = -30$

1. $4x - 8y = 15$

$-5x + 10y = -30$

Explanation:

Problem 15

Step1: Eliminate $x$ via scaling

Multiply second equation by 2: $2(x + 5y) = 2(8) \implies 2x + 10y = 16$

Step2: Subtract first equation

$(2x + 10y) - (2x + 3y) = 16 - 9 \implies 7y = 7$

Step3: Solve for $y$

$y = \frac{7}{7} = 1$

Step4: Substitute $y=1$ to find $x$

$x + 5(1) = 8 \implies x = 8 - 5 = 3$

Step1: Eliminate $y$ via scaling

Multiply first equation by 2: $2(3x + y) = 2(5) \implies 6x + 2y = 10$

Step2: Add to second equation

$(6x + 2y) + (2x - 2y) = 10 + (-2) \implies 8x = 8$

Step3: Solve for $x$

$x = \frac{8}{8} = 1$

Step4: Substitute $x=1$ to find $y$

$3(1) + y = 5 \implies y = 5 - 3 = 2$

Step1: Eliminate $x$ via scaling

Multiply second equation by 2: $2(-3x + 3y) = 2(-6) \implies -6x + 6y = -12$

Step2: Add to first equation

$(6x + 4y) + (-6x + 6y) = 42 + (-12) \implies 10y = 30$

Step3: Solve for $y$

$y = \frac{30}{10} = 3$

Step4: Substitute $y=3$ to find $x$

$6x + 4(3) = 42 \implies 6x = 42 - 12 = 30 \implies x = 5$

Answer:

$x=3, y=1$

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Problem 16