|
Question 391159: Please help me solve these few problems:
15x-4+6x-9=?
5a to the second power b - 1ab to the second power - 7a to the second power b to the second power= -2a to the second power b to the second power - 1ab to the second power=?
3(1x+2)+ 5(1x+3)=8x+21
3(2x-5)- 4(5x-2)=?
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! First:
15x-4+6x-9
Since 15x and 6x are like terms, add 6x to 15x to get 21x.
21x-4-9
Subtract 9 from -4 to get -13.
21x-13
second problem, first you have to solve for A:
5a^(2)b-1ab^(2)-7a^(2)b^(2)=-2a^(2)b^(2)-1ab^(2)
Move all terms not containing a to the right-hand side of the equation.
-7a^(2)b^(2)+5a^(2)b-ab^(2)+2a^(2)b^(2)+ab^(2)=0
Since -7a^(2)b^(2) and 2a^(2)b^(2) are like terms, subtract 2a^(2)b^(2) from -7a^(2)b^(2) to get -5a^(2)b^(2).
-5a^(2)b^(2)+5a^(2)b-ab^(2)+ab^(2)=0
Since -ab^(2) and ab^(2) are like terms, subtract ab^(2) from -ab^(2) to get 0.
-5a^(2)b^(2)+5a^(2)b=0
Factor out the GCF of -5a^(2)b from each term in the polynomial.
-5a^(2)b(b)-5a^(2)b(-1)=0
Factor out the GCF of -5a^(2)b from -5a^(2)b^(2)+5a^(2)b.
-5a^(2)b(b-1)=0
If any individual factor on the left-hand side of the equation is equal to 0, the entire expression will be equal to 0.
-5a^(2)b=0_(b-1)=0
Set the first factor equal to 0 and solve.
-5a^(2)b=0
Divide each term in the equation by -5b.
-(5a^(2)b)/(-5b)=(0)/(-5b)
Simplify the left-hand side of the equation by canceling the common factors.
a^(2)=(0)/(-5b)
Any expression with zero in the numerator is zero.
a^(2)=0
Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.
a=\~(0)
Pull all perfect square roots out from under the radical. In this case, remove the 0 because it is a perfect square.
a=\0
\0 is equal to 0.
a=0
Set the next factor equal to 0 and solve.
(b-1)=0
Remove the parentheses around the expression b-1.
b-1=0
Since -1 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 1 to both sides.
b=1
The final solution is all the values that make -5a^(2)b(b-1)=0 true.
a=0, b=1
Now (just to show you the proper work, solve for B:
\= +- and ~= square root of
5a^(2)b-1ab^(2)-7a^(2)b^(2)=-2a^(2)b^(2)-1ab^(2)
Move all terms not containing b to the right-hand side of the equation.
-7a^(2)b^(2)+5a^(2)b-ab^(2)+2a^(2)b^(2)+ab^(2)=0
Since -7a^(2)b^(2) and 2a^(2)b^(2) are like terms, subtract 2a^(2)b^(2) from -7a^(2)b^(2) to get -5a^(2)b^(2).
-5a^(2)b^(2)+5a^(2)b-ab^(2)+ab^(2)=0
Since -ab^(2) and ab^(2) are like terms, subtract ab^(2) from -ab^(2) to get 0.
-5a^(2)b^(2)+5a^(2)b=0
Multiply each term in the equation by -1.
-5a^(2)b^(2)*-1+5a^(2)b*-1=0*-1
Simplify the left-hand side of the equation by multiplying out all the terms.
5a^(2)b^(2)-5a^(2)b=0*-1
Multiply 0 by -1 to get 0.
5a^(2)b^(2)-5a^(2)b=0
Use the quadratic formula to find the solutions. In this case, the values are a=5a^(2), b=-5a^(2), and c=0.
b=(-b\~(b^(2)-4ac))/(2a) where ab^(2)+bb+c=0
Use the standard form of the equation to find a, b, and c for this quadratic.
a=5a^(2), b=-5a^(2), and c=0
Substitute in the values of a=5a^(2), b=-5a^(2), and c=0.
b=(-(-5a^(2))\~((-5a^(2))^(2)-4(5a^(2))(0)))/(2(5a^(2)))
Multiply -1 by each term inside the parentheses.
b=(5a^(2)\~((-5a^(2))^(2)-4(5a^(2))(0)))/(2(5a^(2)))
Simplify the section inside the radical.
b=(5a^(2)\5a^(2))/(2(5a^(2)))
Simplify the denominator of the quadratic formula.
b=(5a^(2)\5a^(2))/(10a^(2))
First, solve the + portion of \.
b=(5a^(2)+5a^(2))/(10a^(2))
Simplify the expression to solve for the + portion of the \.
b=1
Next, solve the - portion of \.
b=(5a^(2)-5a^(2))/(10a^(2))
Simplify the expression to solve for the - portion of the \.
b=0
The final answer is the combination of both solutions.
b=1,0
Third Problem:
3(1x+2)+5(1x+3)=8x+21
Arrange the variables alphabetically within the expression 1x. This is the standard way of writing an expression.
3(x+2)+5(1x+3)=8x+21
Arrange the variables alphabetically within the expression 1x. This is the standard way of writing an expression.
3(x+2)+5(x+3)=8x+21
Multiply 3 by each term inside the parentheses.
3x+6+5(x+3)=8x+21
Multiply 5 by each term inside the parentheses.
3x+6+5x+15=8x+21
Since 3x and 5x are like terms, add 5x to 3x to get 8x.
8x+6+15=8x+21
Add 15 to 6 to get 21.
8x+21=8x+21
Since 8x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 8x from both sides.
8x+21-8x=21
Since 8x and -8x are like terms, add -8x to 8x to get 0.
0+21=21
Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.
21=21
Since 21=21, the equation will always be true.
Always True
Last:
All you can do with this, like the first is simplyfy because there is nothing to solve for on the opposite side of the equal mark
3(2x-5)-4(5x-2)
Multiply 3 by each term inside the parentheses.
6x-15-4(5x-2)
Multiply -4 by each term inside the parentheses.
6x-15-20x+8
Since 6x and -20x are like terms, add -20x to 6x to get -14x.
-14x-15+8
Add 8 to -15 to get -7.
-14x-7
|
|
|
| |
