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I moved your question to the trash. It doesn't make any sense. If you key in your algebra problem correctly, I will help you solve it.
You need to re-write your question with the correct notation. Do you mean this:
(-6y)-(12y)*(7-4y)? The way you wrote it out doesnt make sense.
is it y to the power of 4 or 4y.? doesn't make any sense..
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Solving -2x 3y14 and x 2y7 with substitution?
-2x+3y+14 and x+2y+7 1. x+2y+7=0 2. x= -2y-7 3. (substitute new equation in for x) -2(-2y-7)+3y+14=0 4. (distribute) 4y+14+3y+14=0 5. (combine like terms) 7y+28=0 6. (subtract 28) 7y= -28 7. (28 divided by 7 to solve for y) y= -4 8. (substitute -4 value for y) x+2(-4)+7 9. (solve for x value) x-8+7 x-1=0 x=1 10. (answer check) -2(1)+3(-4)+14= 0 (1)+2(-4)+7=0
Solve the following system of equations x plus 2y -6 equals z 3y-2z equals 7 4 plus 3x equals 2y -5z?
x+2y-6=z -z -z x+2y-z-6=0 +6 +6 ---------> x+2y-z=6 3y-2z=7 ---------> 0x+3y-2z=7 4+3x=2y-5z -3x -3x ---------> -3x+2y-5z=4 Put them into a matrix, for x,y,z and their answers. Solve for [A]-1[B], and the answer comes to: x= 1.75, y= 1.5, and z= -1.25
What is 9y plus 4 equals 2y plus 25?
9Y + 4 = 2Y + 25 subtract 2Y from both sides 7Y + 4 = 25 subtract 4 from both sides 7Y = 21 divide both sides integers by 7 (7/7)Y = 21/7 Y = 3 -----------------check in original equation 9(3) + 4 = 2(3) + 25 27 + 4 = 6 + 25 31 = 31 ---------------checks
How do you solve algebraically the system of equations?
You can solve a system of equations using a few methods: elemination and subsitution. In the elemination method you would multiply each of your equations by a factor that would cause one of the variables to be eleminated. ex: If you had 2x-3y=5 and 3x+6y=4 you would multiply the first equation by a factor of 2 (distributing the 2 to each term), giving you a new equation of 4x-6y=10. Now your y terms would cancel. You line up the equations like: 3x+6y=4 4x-6y=10 you simply cancel the y terms and add the others giving you: 7x=14 you then solve for x (in this case divide by 7) x=2 now you plug in x=2 into one of the first equations and solve 3x+6y=4 3(2)+6y=4 plug in x 6+6y=4 multiply 6y=-2 add like terms y=-1/3 solve for y To solve using the subsitution method you would take an equation and solve for one variable then plug that into the other equation. ex. (using the same equations as before) 2x-3y=5 2x=3y+5 x=(3/2)y+(5/2) 3x+6y=4 3((3/2)y+(5/2))+6y=4 plug in for x (9/2)y+(15/2)+6y=4 distribute the 3 (21/2)y=(-7/2) add like terms y=(-1/3) solve for y you would then plug in y and solve for x 2x-3(-1/3)=5 2x+1=5 2x=4 x=2 Hope this helps!
What type of lines are these x plus 2y equals 7 and 2x - y equals 7?
x + 2y = 7 2y = -x + 7 y = -(x-7)/2 => -x/2 + 7/2 2x - y = 7 -y = -2x + 7 y = 2x + 7 Since -1/2 is the negative reciporical of 2, the slopes of these equations are perpendicular. Therefore, these two lines are perpendicular.
