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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!
Point A' 3 5 is the reflection of point A -1 3 What is the point of reflection?
(1,4) Let (x,y) represent the point of reflection. We know that A and A' must be on the same line as (x,y) and must be equidistant from (x,y). Because they are equidistant from (x,y), we know: (x-3)2+(y-(5))2=(x-(-1))2+(y-(3))2 Now find the line that they are on by using the points A and A': 5=m(3)+b 3=m(-1)+b Subtracting the left and right hand sides of the two equations, gives us: 2=4m -->m=1/2 Using this value of m and substituting into either equation gives us b=7/2. So, all three points are on the line: y=1/2*x+7/2 This is equivalent to x=2y-7. I will substitute this into the first equation to find our point [I could have used y=1/2*x+7/2, but I'm using the value of x to avoid the fractions.] ((2y-7)-3)2+(y-(5))2=((2y-7)-(-1))2+(y-(3))2 (2y-10)2+(y-5)2=(2y-6)2+(y-3)2 4y2-40y+100+y2-10y+25=4y2-24y+36+y2-6y+9 -50y+125=-30y+45 20y=80 y=4 Substituting y=4 into x=2y-7, gives x=1.
