# Thread: Hey guys Algebra 2 and strugling need help with these!!

1. ## Hey guys Algebra 2 and strugling need help with these!!

Equation 1. 2x+2y=6
y=3x+11
Equation 2. 2x-6y=8
y= -1/3x+2
Equation 3. 6x+8y=4
3x-5y=11
Equation 4. 3x+5y=-18
4x-2y=2
Equation 5. 3x-2y+2z=5
x+2y-z=2
4x-y+3z=11
Equation 6. slope= -3 through (4,-2) in slope intercept form

Equation 7.Perpendicular to -2x+3y=12 through (-4,-5) in standard form

Equation 8. x intercept of 6 through (-2,5) in slope intercept form

Can anyone show me how to do these or atleast do one of each type like do number 1 number 4 number 6-8

These are linear systems. You can solve them with two methods : substitution and elimination. I hate the last one so I am going to explain substitution. It always works for linear systems and is quite easy and nice.

$2x+2y=6$
$y=3x+11$

In the second equation, you know that $y = 3x + 11$. Therefore, in the first equation, you can change $y$ to $3x + 11$. Which gives :

$2x + 2(3x + 11) = 6$
$2x + 6x + 22 = 6$
$8x = -16$
$x = -2$

You now know x, substitute it back into the second equation to find y :

$y = 3x + 11$
$y = 3(-2) + 11$
$y = -6 + 11$
$y = 5$

Therefore the solutions are $S = (-2 ; 5)$ Nice, eh ? (the brackets should be braces but I can't get them on the LaTex editor).
Sometimes you will have to work a bit an equation so that one unknown is alone on a side of the equation (y = ...x... or x = ...y...) so that you can substitute easily.

The three last systems, read your lessons on line equations and it should be easy to write a system of two equations and then solve it by substitution/elimination (the last one consists of adding or substracting the two equations to get rid of one unknown, but it doesn't always work).