It is for algebra 1

## solve the linear system using substitution or elimination 5x - 5y=-3

# 3 Answers

Dasan..hi...this is J. Let's address the elimination process first...to do the elimination process with linear equations, there should be another linear equation, with x and y variables to either add or subtract from one another to eliminate either variable...here's an example. (the key to solving these kinds of problems is similar to adding or subtracting fractions.) Short and sweet... take your original equation 1st equation: 5x - 5y = -3 say there was another equation 2nd Equation: 3x - 2y = 7 by least common fractor and adding and subtracting either the x or y values have to equal zero. The least common factor between the two equations is 10y, so how do we do this? Multiply the 1st equation by -2 and the 2nd equation by 5. will look like this -2(5x -5y = -3) and 5( 3x - 2y = 7), complete the multiplication this then results in: -10x +10y = 6 (did you get the multiplication? if not let me know). And for the 2nd equation it looks like this: 15x - 10y = 35(haven't lost you, have I?) Now, do you see that the y-values are both 10, one positive and one negative. So, if we add the two together, it will be like subtracting 10 minus 10. When you do this, place one equation of top of one another, with the y-values directly above the other. It should look like this: (-10 x + 10y = 6) + (15x -10y =35) COMBINE LIKE TERMS: Remember, we are adding: (-10x + 15x) + ( 10y + -10y) = (6 + 35) Can you see that? It is much more easier to see in you place the equations like when adding in columns. Completing the operations: (5x) + (0y) = 41 See that...change the addition sign because of the negative, RIGHT? This then leads to 5x = 41 divide both sides by 5 then x = 41/5 or 41 ÷ 5 now we have a value for x then substitute that value in to the 1st and 2nd equations and solve for y. Hope it isn't too complicated...a little(A LITTLE!!!!!) wordy...it would be great to explain it to you in-person. So, that Dasan is the ELIMINATION PROCESS...but you should have 2 equations. Now...tut tut tah dahhhhh!!! Solve for either variable by addition or subtracting to get either x = something or y = something, then substitute. I'll do x, then you do y, okay? Step One: get x value on one side of the equation using addition of 5y 5x - 5y + 5y = 5y - 3 See that? What you do on one side, you have to do one the other side of the equation. Step 2: complete the operation 5x + 0 = 5y - 3 Step 3: Rewrite the equation 5x = 5y - 3 Step 4: Divide both sides by 5 5x ÷ 5 = (5y - 3) ÷ 5 Step 5: 5 is eliminated on the left side now we have a value for x x = (5y ÷ 5) - (3 ÷ 5) Step 6: Rewrite the equation x = y - 3/5 Now, substitute the x value y - 3/5 into the original equation and solve for y Example x = y - 3/5 5x - 5y = -3 Step 1: 5(y - 3/5) - 5y = -3 Step 2: Complete the operations (5y - 15/5) - 5y = -3 Step 3:Solve fraction (5y - 3) - 5y = -3 Step 4: Combine like-terms 5y - 5y -3 = -3 which turns out to be -3 = -3 Since this is a true statement, the x = y - 3/5 Now solve for y doing the same method...and if equal like the value for x then you will have a x value and a y value. sorry its so long...shortcuts sometimes can confuse...but the more you do these kinds of problems....then you will be able to eliminate steps because you know what operations you are doing to solve the problem. Good luck!

To solve for n variables, you will need to have n equations. Since you have two variables, x and y, you will need two equations to solve the system.

What is the other equation to the system?