Question

solution
$x = \underline{\quad\quad}$, $y = \underline{\quad\quad}$ solution
determine what operation(s) you would use to solve each system by elimination.

1. $3x + 2y = 1$

$4x + 3y = -2$
addition
subtraction
multiply 1 equation/add
multiply 1 equation/subtract
multiply both equations/add
multiply both equations/subtract

1. $2x + 5y = 17$

$6x - 5y = -9$
addition
subtraction
multiply 1 equation/add
multiply 1 equation/subtract
multiply both equations/add
multiply both equations/subtract

1. $-2x + 15y = -32$

$7x - 5y = 17$
addition
subtraction
multiply 1 equation/add
multiply 1 equation/subtract
multiply both equations/add
multiply both equations/subtract
solving a system using elimination by addition/subtraction/multiplication

1. $2x + 5y = 17$

$6x - 5y = -9$
$x = \underline{\quad\quad}$, $y = \underline{\quad\quad}$ solution

1. $7x - 5y = 17$

$-2x + 15y = -32$
$x = \underline{\quad\quad}$, $y = \underline{\quad\quad}$ solution

1. $-3x - 3y = 9$

$3x - 4y = 5$
$x = \underline{\quad\quad}$, $y = \underline{\quad\quad}$ solution

1. $2x + 6y = 18$

$x + 3y = 9$
$x = \underline{\quad\quad}$, $y = \underline{\quad\quad}$ solution

Explanation:

Step1: Solve system 3

Step1.1: Eliminate x, multiply eq1 by 4, eq2 by 3

$4\times(3x+2y=1) \implies 12x+8y=4$
$3\times(4x+3y=-2) \implies 12x+9y=-6$

Step1.2: Subtract new eq1 from new eq2

$(12x+9y)-(12x+8y)=-6-4$
$y=-10$

Step1.3: Substitute y=-10 into eq1

$3x+2(-10)=1 \implies 3x-20=1 \implies 3x=21 \implies x=7$

Step2: Solve system 4

Step2.1: Eliminate y via addition

$(2x+5y)+(6x-5y)=17+(-9)$
$8x=8 \implies x=1$

Step2.2: Substitute x=1 into eq1

$2(1)+5y=17 \implies 2+5y=17 \implies 5y=15 \implies y=3$

Step3: Solve system 5

Step3.1: Eliminate x via addition

$(-2x+15y)+(7x-5y)=-32+17$
$5x+10y=-15 \implies x+2y=-3$

Step3.2: Isolate x, substitute into eq2

$x=-3-2y$
$7(-3-2y)-5y=17 \implies -21-14y-5y=17 \implies -19y=38 \implies y=-2$

Step3.3: Substitute y=-2 into x expression

$x=-3-2(-2)=-3+4=1$

Step4: Solve system 6

Step4.1: Eliminate y via addition

$(2x+5y)+(6x-5y)=17+(-9)$
$8x=8 \implies x=1$

Step4.2: Substitute x=1 into eq1

$2(1)+5y=17 \implies 5y=15 \implies y=3$

Step5: Solve system7

Step5.1: Eliminate x via addition

$(7x-5y)+(-2x+15y)=17+(-32)$
$5x+10y=-15 \implies x+2y=-3$

Step5.2: Isolate x, substitute into eq1

$x=-3-2y$
$7(-3-2y)-5y=17 \implies -21-19y=17 \implies -19y=38 \implies y=-2$

Step5.3: Substitute y=-2 into x expression

$x=-3-2(-2)=1$

Step6: Solve system8

Step6.1: Eliminate x via subtraction

$(-3x-3y)-(3x-4y)=9-5$
$-6x+y=4$

Step6.2: Isolate y, substitute into eq1

$y=6x+4$
$-3x-3(6x+4)=9 \implies -3x-18x-12=9 \implies -21x=21 \implies x=-1$

Step6.3: Substitute x=-1 into y expression

$y=6(-1)+4=-2$

Step7: Solve system9

Step7.1: Eliminate x, multiply eq2 by 2

$2\times(x+3y=9) \implies 2x+6y=18$

Step7.2: Subtract new eq2 from eq1

$(2x+6y)-(2x+6y)=18-18 \implies 0=0$
Infinite solutions: $x=9-3y$ for all real y

Answer:

1. $x=7$, $y=-10$
2. $x=1$, $y=3$
3. $x=1$, $y=-2$
4. $x=1$, $y=3$
5. $x=1$, $y=-2$
6. $x=-1$, $y=-2$
7. Infinite solutions: $x=9-3y$ (for any real number $y$)