Guide :

add or subtract this expression

Square root 6a    -  4square root 54a   -   4 square root 216a simplify by combining like radical terms. Assume all veriables and radicands represent nonnegative numbers

Research, Knowledge and Information :


Adding and Subtracting Rational Expressions - Purplemath


Demonstrates the steps involved in adding rational expressions, comparing polynomial fractions to regular fractions. ... "Adding and Subtracting Rational Expressions."
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Adding and Subtracting Expressions - Online Math Learning


Adding and Subtracting Algebraic Expressions, ... In these lessons, we will learn how to simplify expressions by adding or subtracting like terms ...
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Subtracting rational expressions (video) | Khan Academy


Practice: Add & subtract rational expressions: factored denominators. Subtracting rational expressions. Practice: Add & subtract rational expressions. Next tutorial.
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Intro to adding rational expressions with unlike denominators ...


Intro to adding rational expressions with unlike denominators. ... Add & subtract rational expressions: like denominators. Adding rational expression: unlike ...
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1. Addition and Subtraction of Algebraic Expressions


A term in an algebraic expression is an expression involving letters and/or numbers (called factors), multiplied together. Important: We can only add or subtract like ...
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Adding and Subtracting Rational Expressions - YouTube


Mar 26, 2008 · Adding and Subtracting Rational Expressions! (NOT EQUATIONS!, BUT EXPRESSIONS!) ... Rational Expressions , Adding, Subtracting, Multiplying, Dividing, ...

Adding & Subtracting Rational Expressions


ADDING AND SUBTRACTING RATIONAL EXPRESSIONSADDING AND SUBTRACTING RATIONAL EXPRESSIONS To Add or Subtract ... or subtracting rational expressions subtracting ...
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7.4 Rational Expressions - Add & Subtract


Rational Expressions - Add & Subtract Objective: ... Add or subtract the rational expressions. Simplify your answers whenever possible. 1) 2 a+3 + 4 a +3 3) t 2 +4t t ...
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Wolfram|Alpha Widgets: "Add & Sub Rational Expressions ...


Add & Sub Rational Expressions. Added Oct 25, 2012 by elecKctra in Mathematics. Add & Sub Rational Expressions. Send feedback ... To add a widget to a MediaWiki site, ...
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Adding and Subtracting and Simplifying Linear Expressions (A)


The Adding and Subtracting and ... The Adding and Subtracting and Simplifying Linear Expressions (A) ... algebra, linear, expression, simplify, add, subtract: Size of ...
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Suggested Questions And Answer :


sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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( 3x - 2 ) ( x - 1) - ( 2x + 5 ) ( 5x + 6 )

(3x-2)(x-1)-(2x+5)(5x+6) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. (3x*x+3x*-1-2*x-2*-1)-(2x+5)(5x+6) Simplify the FOIL expression by multiplying and combining all like terms. (3x^(2)-5x+2)-(2x+5)(5x+6) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. (3x^(2)-5x+2)-(2x*5x+2x*6+5*5x+5*6) Simplify the FOIL expression by multiplying and combining all like terms. (3x^(2)-5x+2)-(10x^(2)+37x+30) Multiply -1 by each term inside the parentheses. (3x^(2)-5x+2)+(-10x^(2)-37x-30) Remove the parentheses that are not needed from the expression. 3x^(2)-5x+2-10x^(2)-37x-30 Since 3x^(2) and -10x^(2) are like terms, add -10x^(2) to 3x^(2) to get -7x^(2). -7x^(2)-5x+2-37x-30 Since -5x and -37x are like terms, subtract 37x from -5x to get -42x. -7x^(2)-42x+2-30 Subtract 30 from 2 to get -28. -7x^(2)-42x-28
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find the value of ”x”.

130+110+x+(x-3)=360 Remove the parentheses around the expression x-3. 130+110+x+x-3=360 Add 110 to 130 to get 240. 240+x+x-3=360 Subtract 3 from 240 to get 237. 237+x+x=360 Since x and x are like terms, add x to x to get 2x. 237+2x=360 Since 237 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 237 from both sides. 2x=-237+360 Add 360 to -237 to get 123. 2x=123 Divide each term in the equation by 2. (2x)/(2)=(123)/(2) Simplify the left-hand side of the equation by canceling the common terms. x=(123)/(2)
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Which of the following operations will isolate the variable in te expression 8x-9?

mi ges is that stuf shood be 8x-9=0 . . . 8x=9 . . . x=9/8 . . . x=1.125
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5(2x + 1) = 3(x + 4) -5

5(2x+1)=3(x+4)-5 Multiply 3 by each term inside the parentheses. 5(2x+1)=(3x+12)-5 Subtract 5 from 12 to get 7. 5(2x+1)=3x+7 Multiply 5 by each term inside the parentheses. (10x+5)=3x+7 Remove the parentheses around the expression 10x+5. 10x+5=3x+7 Since 3x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 3x from both sides. 10x+5-3x=7 Since 10x and -3x are like terms, add -3x to 10x to get 7x. 7x+5=7 Since 5 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 5 from both sides. 7x=-5+7 Add 7 to -5 to get 2. 7x=2 Divide each term in the equation by 7. (7x)/(7)=(2)/(7) Simplify the left-hand side of the equation by canceling the common terms. x=(2)/(7)
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130 + 110 + x + (x - 3) = 360

130+110+x+(x-3)=360 Remove the parentheses around the expression x-3. 130+110+x+x-3=360 Add 110 to 130 to get 240. 240+x+x-3=360 Subtract 3 from 240 to get 237. 237+x+x=360 Since x and x are like terms, add x to x to get 2x. 237+2x=360 Since 237 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 237 from both sides. 2x=-237+360 Add 360 to -237 to get 123. 2x=123 Divide each term in the equation by 2. (2x)/(2)=(123)/(2) Simplify the left-hand side of the equation by canceling the common terms. x=(123)/(2)
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solve equation 6x+5y=0, 3x-2y = 19

Exponents Exponents are supported on variables using the ^ (caret) symbol. For example, to express x2, enter x^2. Note: exponents must be positive integers, no negatives, decimals, or variables. Exponents may not currently be placed on numbers, brackets, or parentheses. Parentheses and Brackets Parentheses ( ) and brackets [ ] may be used to group terms as in a standard equation or expression. Multiplication, Addition, and Subtraction For addition and subtraction, use the standard + and - symbols respectively. For multiplication, use the * symbol. A * symbol is not necessiary when multiplying a number by a variable. For instance: 2 * x can also be entered as 2x. Similarly, 2 * (x + 5) can also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5). The * is also optional when multiplying with parentheses, example: (x + 1)(x - 1). Order of Operations The calculator follows the standard order of operations taught by most algebra books - Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. The only exception is that division is not currently supported; attempts to use the / symbol will result in an error. Division, Square Root, Radicals, Fractions The above features are not supported at this time. A future release will add this functionality.
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add or subtract the linear expression (4x-10)-(3x+7)

(4x-10) -(3x+7) =4x-3x -10-7 =x-17
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can you please show me how to solve -7(x+7)+68=2x-9(x+1)

-7(x+7)+68=2x-9(x+1) Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2x-9(x+1)=-7(x+7)+68 Multiply -7 by each term inside the parentheses. 2x-9(x+1)=(-7x-49)+68 Add 68 to -49 to get 19. 2x-9(x+1)=-7x+19 Multiply -9 by each term inside the parentheses. 2x+(-9x-9)=-7x+19 Since 2x and -9x are like terms, add -9x to 2x to get -7x. -7x-9=-7x+19 Since -7x contains the variable to solve for, move it to the left-hand side of the equation by adding 7x to both sides. -7x-9+7x=19 Since -7x and 7x are like terms, subtract 7x from -7x to get 0. 0-9=19 Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression. -9=19 Since -9/=19, there are no solutions. No Solution
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