Question

solving systems of equations: elimination
the system below would not be too fun to solve by substitution. follow along to learn a new trick...
substitution
isolate y in the first equation: 5y=-6x + 17
y = -\frac{6}{5}x+\frac{17}{5}
substitute: 9x - 5(-\frac{6}{5}x+\frac{17}{5}) = 13
9x + 6x-17 = 13
15x-17 = 13
15x = 30
x=
solve for y: y=-\frac{6}{5}(2)+\frac{17}{5}
y=
solution:
elimination
6x + 5y = 17
9x - 5y = 13
solution:
the elimination method:

1. create like terms!
2. the equations together, thus eliminating one of the variable terms!
3. solve this one - step equation.
4. use the variable you know to find the one you dont!

lets solve the following systems of equations by elimination.
-2x + y = 14
2x + 8y = 22
solution:
-x + 2y=-13
-x - 2y = 11
solution:

Explanation:

Step1: Solve the first system by substitution

• Isolate $y$ in $6x + 5y=17$: $y=-\frac{6}{5}x+\frac{17}{5}$
• Substitute into $9x - 5y = 13$:
• $9x-5(-\frac{6}{5}x+\frac{17}{5})=13$
• $9x + 6x-17 = 13$
• $15x=30$, so $x = 2$
• Solve for $y$: $y=-\frac{6}{5}(2)+\frac{17}{5}=\frac{-12 + 17}{5}=1$

Step2: Fill in the blanks for the elimination - method steps

1. Opposite; we create opposite - like terms (e.g., $5y$ and $- 5y$ in the first system)
2. Add; we add the two equations together to eliminate a variable

Step3: Solve the second system $-2x + y=14$ and $2x+8y = 22$ by elimination

• Add the two equations: $(-2x + y)+(2x + 8y)=14 + 22$
• $9y=36$, so $y = 4$
• Substitute $y = 4$ into $-2x + y=14$: $-2x+4 = 14$
• $-2x=10$, so $x=-5$

Step4: Solve the third system $-x + 2y=-13$ and $-x-2y = 11$ by elimination

• Subtract the second equation from the first: $(-x + 2y)-(-x-2y)=-13 - 11$
• $4y=-24$, so $y=-6$
• Substitute $y=-6$ into $-x + 2y=-13$: $-x+2(-6)=-13$
• $-x-12=-13$
• $-x=-1$, so $x = 1$

Answer:

• In the first substitution - method part: $x = 2$, $y = 1$
• Blanks in elimination - method: 1. Opposite; 2. Add
• Solution for $-2x + y=14$ and $2x+8y = 22$: $x=-5$, $y = 4$
• Solution for $-x + 2y=-13$ and $-x-2y = 11$: $x = 1$, $y=-6$