Question

ip: 7.3 solving a system by elimination - day 1 page #
solve the system using elimination. you should be elimination by addition/subtration only

1. $x + 3y = 5$

$-x - y = -3$
$x = $__, $y = $ solution __

1. $4x + 3y = -5$

$-x + 3y = -10$
$x = $__, $y = $ solution __

1. $2x + 5y = 16$

$3x - 5y = -1$
$x = $__, $y = $ solution __

1. $5x + 2y = 18$

$-5x + 3y = -23$
$x = $__, $y = $ solution __

1. describe and correct the error.

description of error: ____
correction: ____
(image: $5x + 2y = 9$ (equation 1)
$3x - 2y = -1$ (equation 2)
$overline{2x = 10}$
$x = 5$
the solution is $(5, -8)$.)
solve the system using elimination. you should be elimination by multiplication.

1. $2x - y = 0$

$3x - 2y = -3$
$x = $__, $y = $ solution __

1. $10x + 16y = -4$

$-2x + 3y = 7$
$x = $__, $y = $ solution __

Explanation:

Problem 2

Step1: Subtract equations to eliminate $y$

$(4x + 3y) - (-x + 3y) = -5 - (-10)$
$4x + 3y + x - 3y = 5$
$5x = 5$

Step2: Solve for $x$

$x = \frac{5}{5} = 1$

Step3: Substitute $x=1$ to find $y$

$4(1) + 3y = -5$
$4 + 3y = -5$
$3y = -5 - 4 = -9$
$y = \frac{-9}{3} = -3$

Problem 3

Step1: Add equations to eliminate $y$

$(2x + 5y) + (3x - 5y) = 16 + (-1)$
$2x + 5y + 3x - 5y = 15$
$5x = 15$

Step2: Solve for $x$

$x = \frac{15}{5} = 3$

Step3: Substitute $x=3$ to find $y$

$2(3) + 5y = 16$
$6 + 5y = 16$
$5y = 16 - 6 = 10$
$y = \frac{10}{5} = 2$

Problem 4

Step1: Add equations to eliminate $x$

$(5x + 2y) + (-5x + 3y) = 18 + (-23)$
$5x + 2y -5x + 3y = -5$
$5y = -5$

Step2: Solve for $y$

$y = \frac{-5}{5} = -1$

Step3: Substitute $y=-1$ to find $x$

$5x + 2(-1) = 18$
$5x - 2 = 18$
$5x = 18 + 2 = 20$
$x = \frac{20}{5} = 4$

Problem 5

Step1: Identify addition error

When adding $9 + (-1)$, the result was incorrectly calculated as 10 instead of 8.

Step2: Correct $x$ calculation

$2x = 9 + (-1) = 8$
$x = \frac{8}{2} = 4$

Step3: Substitute $x=4$ to find $y$

$5(4) + 2y = 9$
$20 + 2y = 9$
$2y = 9 - 20 = -11$
$y = \frac{-11}{2} = -5.5$

Problem 9

Step1: Multiply first equation by 2

$2(2x - y) = 2(0)$
$4x - 2y = 0$

Step2: Subtract second equation to eliminate $y$

$(4x - 2y) - (3x - 2y) = 0 - (-3)$
$4x - 2y -3x + 2y = 3$
$x = 3$

Step3: Substitute $x=3$ to find $y$

$2(3) - y = 0$
$6 - y = 0$
$y = 6$

Problem 10

Step1: Multiply second equation by 5

$5(-2x + 3y) = 5(7)$
$-10x + 15y = 35$

Step2: Add to first equation to eliminate $x$

$(10x + 16y) + (-10x + 15y) = -4 + 35$
$10x +16y -10x +15y = 31$
$31y = 31$

Step3: Solve for $y$

$y = \frac{31}{31} = 1$

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

$10x + 16(1) = -4$
$10x +16 = -4$
$10x = -4 -16 = -20$
$x = \frac{-20}{10} = -2$

Answer:

1. $x=-1$, $y=2$, Solution: $(-1, 2)$ (corrected from original work error: subtracting $x+3y=5$ and $x-y=-3$ gives $4y=8$, so $y=2$, then $x=5-3(2)=-1$)
2. $x=1$, $y=-3$, Solution: $(1, -3)$
3. $x=3$, $y=2$, Solution: $(3, 2)$
4. $x=4$, $y=-1$, Solution: $(4, -1)$
5. Description of Error: Incorrect addition of constants when eliminating $y$; $9 + (-1)$ was calculated as 10 instead of 8.

Correction: $2x=8$, so $x=4$. Substitute $x=4$ into $5x+2y=9$: $20+2y=9$, $2y=-11$, $y=-5.5$. Solution is $(4, -5.5)$

1. $x=3$, $y=6$, Solution: $(3, 6)$
2. $x=-2$, $y=1$, Solution: $(-2, 1)$