Question

. Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this.
-x + 2y = 8

-x + 10y = 48

Answer 1

subtract the first equation from the second equation to get:
8y = 40

can you take it from here?

Answer 2

Do you know how to use Matrices?

Answer 3

NO

Answer 4

Ok use dplanc's approach.

Answer 5

HOW WOULD I DO

Answer 6

change the signs of each term in the first equation to get x-2y = -8


Then add the equations

x - 2y = -8
-x + 10y = 48
---------------
0x + 8y = 40

So we have 8y = 40

Solve 8y = 40 for y

WHAT WILL I
DO NEXT

Answer 7

-x + 2y = 8 -(1)
-x + 10y = 48 -(2)

(2) -(1)
(-x + 10y) - (-x + 2y) = 48 - 8
-x + 10y + x - 2y = 40
8y = 40
y = ...?

Sub y = ... into (1) to get x.

Answer 8

SO WHERE THERE IS A Y PUY 1

Answer 9

Can you solve y first
8y = 40
y = ...?

Answer 10

Y=5

Answer 11

If you have two equations with two unkowns (x and y) , one way to solve this "system" of equations is to multiply one equation by a constant such that when it is added to the second
one of the variables is eliminated from the resulting equation. for example for the system of equations
\[2x+3y=4 \] with \[x+y=3\] we can multiply the second equation by \[-2\] to get \[-2x-2y=-6\], you are then allowed to add this to the first equation;
\[ (2x+3y)+(-2x-2y) =4-6 \]
gives
\[ 0x+(3-2)y=-2\]
and
\[y=-2\] once you have this substitute it back into equation \[2x+3y=4\] and get \[2x-6=4\] then solveing for x, \[2x=-2\],\[x=-1\].

Answer 12

Use the same idea for your problem.

Answer 13

Or do you need another example?

Answer 14

As you've found, y=5
Now, put y=5 into (1)
-x + 2(5) = 8
Can you find x?