Extended Distributive Property: Matching expressions equal in value

Solving the problemUsing a rectangleDistributive property in geometrySolving an equation using the extended distributive lawUsing additional geometric shapesAscertain whether the law of distributive property is applicableIs the equality correct?Using variablesComplete the missing numberApplying the formula
Question 1

Join expressions of equal value

  1. \( (12x-5)(y+2) \)

  2. \( (x-12)(5y+2) \)

  3. \( (12x+5)(y-2) \)

    a.\( 12xy-24x+5y-10 \)

    b.\( 12xy+24x-5y-10 \)

    c.\( 5xy+2x-60y-24 \)

Question 2

Join expressions of equal value

  1. \( (a-b)(c-4) \)

  2. \( (a+b)(c+4) \)

  3. \( (a-b)(c+4) \)

  4. \( (a+b)(c-4) \)

    a.\( ac-4a+bc-4b \)

    b.\( ac+4a-bc-4b \)

    c.\( ac-4a-bc+4b \)

    d.\( ac+4a+bc+4b \)

Question 3

Join expressions of equal value

  1. \( (a+b)(c+d) \)

  2. \( (a+c)(b+d) \)

  3. \( (a+d)(c+b) \)

    a.\( ac+ad+bc+bd \)

    b.\( ac+ab+dc+db \)

    c.\( ab+ad+cb+cd \)

Question 4

Join expressions of equal value

  1. \( (x+6)(x+8) \)

  2. \( (6+x)(8-x) \)

  3. \( (x+x)(6+8) \)

    a.\( 48+2x-x^2 \)

    b.\( 28x \)

    c.\( x^2+14x+48 \)

Question 5

Join expressions that have the same value

  1. \( (x+4)(x-3) \)

  2. \( (x+4)(x+3) \)

  3. \( (x-4)(x-3) \)

    a.\( x^2+x-12 \)

    b.\( x^2-7x+12 \)

    c.\( x^2+7x+12 \)

Examples with solutions for Extended Distributive Property: Matching expressions equal in value

Exercise #1

Join expressions of equal value

  1. (12x−5)(y+2) (12x-5)(y+2)

  2. (x−12)(5y+2) (x-12)(5y+2)

  3. (12x+5)(y−2) (12x+5)(y-2)

    a.12xy−24x+5y−10 12xy-24x+5y-10

    b.12xy+24x−5y−10 12xy+24x-5y-10

    c.5xy+2x−60y−24 5xy+2x-60y-24

Video Solution

Step-by-Step Solution

Let's simplify the given expressions, open the parentheses using the extended distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition, therefore we won't forget of course that the sign preceding the term is an integral part of it, and we will also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we open using the above formula, first, as an expression where addition operation exists between all terms (if required),

We will therefore simplify each of the expressions in the given problem, while being careful about the above, first opening the parentheses using the aforementioned distribution law and then using the commutative law of addition and multiplication and combining like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (12x−5)(y+2)↓(12x+(−5))(y+2)12x⋅y+12x⋅2+(−5)⋅y+(−5)⋅212xy+24x−5y−10 (12x-5)(y+2) \\ \downarrow\\ \big(12x+(-5)\big)(y+2)\\ 12x\cdot y+12x\cdot 2+(-5)\cdot y+(-5)\cdot 2\\ \boxed{12xy+24x-5y-10}\\

  2. (x−12)(5y+2)↓(x+(−12))(5y+2)x⋅5y+x⋅2+(−12)⋅5y+(−12)⋅25xy+2x−60y−24 (x-12)(5y+2) \\ \downarrow\\ \big(x+(-12)\big)(5y+2)\\ x\cdot 5y+x\cdot 2+(-12)\cdot 5y+(-12)\cdot 2\\ \boxed{5xy+2x-60y-24}\\

  3. (12x+5)(y−2)↓(12x+5)(y+(−2))12x⋅y+12x⋅(−2)+5⋅y+5⋅(−2) 12xy−24x+5y−10 (12x+5)(y-2) \\ \downarrow\\ (12x+5)\big(y+(-2)\big)\\ 12x\cdot y+12x\cdot (-2)+5\cdot y+5\cdot (-2)\\\ \boxed{12xy-24x+5y-10}\\

    As can be noticed, in all expressions where we started the multiplication between the expressions in parentheses above, the result of multiplication (obtained after applying the aforementioned distribution law) yielded an expression where terms cannot be combined, and this is because all terms in the resulting expression are different from each other (remember that all variables in like terms need to be identical and have the same exponent),

    Now, let's use the commutative law of addition and multiplication to observe that:

    The simplified expression in 1 matches the expression in option B,

    The simplified expression in 2 matches the expression in option C,

    The simplified expression in 3 matches the expression in option A,

Therefore, the correct answer (among the suggested options) is answer B.

Answer

1-c, 2-b, 3-a

Exercise #2

Join expressions of equal value

  1. (a−b)(c−4) (a-b)(c-4)

  2. (a+b)(c+4) (a+b)(c+4)

  3. (a−b)(c+4) (a-b)(c+4)

  4. (a+b)(c−4) (a+b)(c-4)

    a.ac−4a+bc−4b ac-4a+bc-4b

    b.ac+4a−bc−4b ac+4a-bc-4b

    c.ac−4a−bc+4b ac-4a-bc+4b

    d.ac+4a+bc+4b ac+4a+bc+4b

Video Solution

Step-by-Step Solution

We use all the exercises of the extended distributive property:(a+b)×(c+d)=ac+ad+bc+bd (a+b)\times(c+d)=ac+ad+bc+bd

1.(a−b)(c−4)=ac−4a−bc+4b (a-b)(c-4)=ac-4a-bc+4b

2.(a+b)(c+4)=ac+4a+bc+4b (a+b)(c+4)=ac+4a+bc+4b

3.(a−b)(c+4)=ac+4a−bc−4b (a-b)(c+4)=ac+4a-bc-4b

4.(a+b)(c−4)=ac−4a+bc−4b (a+b)(c-4)=ac-4a+bc-4b

Answer

1-c, 2-d, 3-b, 4-a

Exercise #3

Join expressions of equal value

  1. (a+b)(c+d) (a+b)(c+d)

  2. (a+c)(b+d) (a+c)(b+d)

  3. (a+d)(c+b) (a+d)(c+b)

    a.ac+ad+bc+bd ac+ad+bc+bd

    b.ac+ab+dc+db ac+ab+dc+db

    c.ab+ad+cb+cd ab+ad+cb+cd

Video Solution

Step-by-Step Solution

Let's simplify the given expressions, open the parentheses using the extended distribution law:

(x+y)(m+r)=xm+xr+ym+yr (\textcolor{red}{x}+\textcolor{blue}{y})(m+r)=\textcolor{red}{x}m+\textcolor{red}{x}r+\textcolor{blue}{y}m+\textcolor{blue}{y}r

Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition, therefore we won't forget of course that the leading sign of the term is an inseparable part of it, and we will also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition operation exists between all terms (if needed),

We will therefore simplify each of the expressions in the given problem, while being careful about the above, first opening the parentheses using the mentioned distribution law and then using the commutative law of addition and multiplication and combining like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (a+b)(c+d)ac+ad+bc+bd (a+b)(c+d) \\ \boxed{ac+ad+bc+bd}\\

  2. (a+c)(b+d)ab+ad+cb+cd (a+c)(b+d) \\ \boxed{ab+ad+cb+cd}

  3. (a+d)(c+b)ac+ab+dc+db (a+d)(c+b) \\ \boxed{ac+ab+dc+db}

    As can be noticed, in all expressions where we started the multiplication between the expressions in parentheses above, the result of multiplication (obtained after applying the mentioned distribution law) yielded an expression where terms cannot be combined, and this is because all terms in the resulting expression are different from each other (we'll remind that all variables in like terms need to be identical and have the same exponent),

    Now, we'll use the commutative law of addition and multiplication to observe that:

    The simplified expression in 1 matches the expression in option A,

    The simplified expression in 2 matches the expression in option C,

    The simplified expression in 3 matches the expression in option B,

Therefore the correct answer (among the suggested options) is answer C.

Answer

1-a, 2-b, 3-b

Exercise #4

Join expressions of equal value

  1. (x+6)(x+8) (x+6)(x+8)

  2. (6+x)(8−x) (6+x)(8-x)

  3. (x+x)(6+8) (x+x)(6+8)

    a.48+2x−x2 48+2x-x^2

    b.28x 28x

    c.x2+14x+48 x^2+14x+48

Video Solution

Step-by-Step Solution

Let's simplify the given expressions, open the parentheses using the extended distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d

Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition, therefore we won't forget of course that the sign preceding the term is an inseparable part of it, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition operation exists between all terms (if required),

We will therefore simply solve each of the expressions in the given problem, while being mindful of the above, first opening the parentheses using the aforementioned distribution law and then using the commutative law of addition and multiplication and combining like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (x+6)(x+8)x⋅x+x⋅8+6⋅x+6⋅8x2+8x+6x+48x2+14x+48 (x+6)(x+8) \\ x\cdot x+x\cdot 8+6\cdot x+6\cdot8\\ x^2+8x+6x+48\\ \boxed{x^2+14x+48}\\

  2. (6+x)(8−x)↓(6+x)(8+(−x))6⋅8+6⋅(−x)+x⋅8+x⋅(−x)48−6x+8x−x248+2x−x2 (6+x)(8-x) \\ \downarrow\\ (6+x)\big(8+(-x)\big) \\ 6\cdot 8+6\cdot (-x)+x\cdot 8+x\cdot(-x)\\ 48-6x+8x-x^2\\ \boxed{48+2x-x^2}\\

  3. (x+x)(6+8)2x⋅1428x (x+x)(6+8) \\ 2x\cdot14\\ \boxed{28x}

    In the last expression we simplified above, we first combined like terms in each of the expressions within parentheses, therefore in this case there was no need to use the extended distribution law mentioned at the beginning of the solution to simplify the expression.

    Now, let's use the commutative law of addition and multiplication to observe that:

    The simplified expression in 1 matches the expression in option C,

    The simplified expression in 2 matches the expression in option A,

    The simplified expression in 3 matches the expression in option B,

Therefore, the correct answer (among the suggested options) is answer B.

Answer

1-b, 2-a, 3-b

Exercise #5

Join expressions that have the same value

  1. (x+4)(x−3) (x+4)(x-3)

  2. (x+4)(x+3) (x+4)(x+3)

  3. (x−4)(x−3) (x-4)(x-3)

    a.x2+x−12 x^2+x-12

    b.x2−7x+12 x^2-7x+12

    c.x2+7x+12 x^2+7x+12

Video Solution

Step-by-Step Solution

Let's simplify the given expressions, open the parentheses using the extended distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an inseparable part of it, and we'll also apply the laws of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition operation exists between all terms (if needed),

We will therefore simplify each of the expressions in the given problem, while being careful about the above, first opening the parentheses using the mentioned distribution law and then using the commutative law of addition and multiplication and combining like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (x+4)(x−3)↓(x+4)(x+(−3))x⋅x+x⋅(−3)+4⋅x+4⋅(−3)x2−3x+4x−12x2+x−12 (x+4)(x-3) \\ \downarrow\\ (x+4)\big(x+(-3)\big) \\ x\cdot x+x\cdot(-3)+4\cdot x+4\cdot(-3)\\ x^2-3x+4x-12\\ \boxed{x^2+x-12}

  2. (x+4)(x+3)x⋅x+x⋅3+4⋅x+4⋅3x2+3x+4x+12x2+7x+12 (x+4)(x+3) \\ x\cdot x+x\cdot 3+4\cdot x+4\cdot 3\\ x^2+3x+4x+12\\ \boxed{x^2+7x+12}

  3. (x−4)(x−3)↓(x+(−4))(x+(−3))x⋅x+x⋅(−3)+(−4)⋅x+(−4)⋅(−3)x2−3x−4x+12x2−7x+12 (x-4)(x-3) \\ \downarrow\\ \big(x+(-4)\big)\big(x+(-3)\big) \\ x\cdot x+x\cdot(-3)+(-4)\cdot x+(-4)\cdot(-3)\\ x^2-3x-4x+12\\ \boxed{x^2-7x+12}

    Now, we'll use the commutative law of addition and multiplication to notice that:

    The simplified expression in 1 matches the expression in option A,

    The simplified expression in 2 matches the expression in option C,

    The simplified expression in 3 matches the expression in option B,

Therefore, the correct answer (among the suggested options) is answer B.

Answer

1-a, 2-c, 3-b

Question 1

Match the expressions (numbers) with the equivalent expressions (letters):

  1. \( (2x-y)(x+3) \)

  2. \( (y-2x)(3-x) \)

  3. \( (2x+y)(x-3) \)

    a.\( 2x^2-6x+yx-3y \)

    b.\( 2x^2-6x-yx+3y \)

    c.\( 2x^2+6x-yx-3y \)

Question 2

Match the expressions that have the same value:

  1. \( (2x+y)(x+2y) \)

  2. \( (2x+2y)(x+y) \)

  3. \( (2x-y)(x-2y) \)

    a. \( 2x^2+4xy+2y^2 \)

    b. \( 2x^2-5xy+2y^2 \)

    c. \( 2x^2+5xy+2y^2 \)

Question 3

Group the expressions that have the same value.

  1. \( (b+c)(a-4) \)

  2. \( (4+c)(a+b) \)

  3. \( (a+4)(b-c) \)

  4. \( (b+4)(c-a) \)

    a. \( ac+ab-4b-4c \)

    b. \( 4b+ab-4c-ac \)

    c. \( bc-ab+4c-4a \)

    d. \( 4a+4b+ac+cb \)

Question 4

Join expressions of equal value

  1. \( (2a+b)(b+4) \)

  2. \( (4+a)(2b+b) \)

  3. \( (2a-b)(b-4) \)

  4. \( (2a-b)(b+4) \)

    a.\( 2ab-8a-b^2+4b \)

    b.\( 12b+3ab \)

    c.\( 2ab+8a-b^2-4b \)

    d.\( 2ab+8a+b^2+4b \)

Question 5

Join expressions of equal value

  1. \( (2x+9)(y+4) \)

  2. \( (2y+9)(x+4) \)

  3. \( (2x-9)(y-4) \)

  4. \( (2x+9)(y-4) \)

    a.\( 2xy+8x+9y+36 \)

    b.\( 2xy-8x+9y-36 \)

    c.\( 2xy-8x-9y+36 \)

    d.\( 2xy+8y+9x+36 \)

Exercise #6

Match the expressions (numbers) with the equivalent expressions (letters):

  1. (2x−y)(x+3) (2x-y)(x+3)

  2. (y−2x)(3−x) (y-2x)(3-x)

  3. (2x+y)(x−3) (2x+y)(x-3)

    a.2x2−6x+yx−3y 2x^2-6x+yx-3y

    b.2x2−6x−yx+3y 2x^2-6x-yx+3y

    c.2x2+6x−yx−3y 2x^2+6x-yx-3y

Video Solution

Step-by-Step Solution

Simplify the given expressions, open parentheses using the extended distributive property:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Keep in mind that in the formula form for the distributive property mentioned above, we assume by default that the operation between the terms inside the parentheses is an addition, therefore, of course, we will not forget that the sign of the term's coefficient is an inseparable part of it. Furthermore, we will apply the rules of sign multiplication and thus we can present any expression within parentheses, which is opened with the help of the previous formula, first, as an expression in which an addition operation takes place among all the terms (if necessary),

Then we will simplify each and every one of the expressions of the given problem, respecting the above, first opening the parentheses through the previously mentioned distributive property. Then we will use the substitution property in addition and multiplication before introducing like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (2x−y)(x+3)↓(2x+(−y))(x+3)2x⋅x+2x⋅3+(−y)⋅x+(−y)⋅32x2+6x−yx−3y (2x-y)(x+3) \\ \downarrow\\ \big(2x+(-y)\big)(x+3) \\ 2x\cdot x+2x\cdot 3+(-y)\cdot x+(-y)\cdot3\\ \boxed{2x^2+6x-yx-3y}\\

  2. (y−2x)(3−x)↓(y+(−2x))(3+(−x))y⋅3+y⋅(−x)+(−2x)⋅3+(−2x)⋅(−x)3y−xy−6x+2x2 (y-2x)(3-x) \\ \downarrow\\ \big(y+(-2x)\big)\big(3+(-x)\big) \\ y\cdot 3+y\cdot (-x)+(-2x)\cdot 3+(-2x)\cdot(-x)\\ \boxed{3y-xy-6x+2x^2}\\

  3. (2x+y)(x−3)↓(2x+y)(x+(−3))2x⋅x+2x⋅(−3)+y⋅x+y⋅(−3)2x2−6x+yx−3y (2x+y)(x-3) \\ \downarrow\\ (2x+y)(x+(-3)) \\ 2x\cdot x+2x\cdot (-3)+y\cdot x+y\cdot(-3)\\ \boxed{2x^2-6x+yx-3y}\\ As you can notice, in all the expressions where we applied multiplication between the expressions in the previous parentheses, the result of the multiplication (obtained after applying the previously mentioned distributive property) produced an expression in which terms cannot be added, and this is because all the terms in the resulting expression are different from each other (remember that all like variables must be identical and in the same power),

    Now, let's use the substitution property in addition and multiplication to distinguish that:

    The simplified expression in 1 corresponds to the expression in option C,

    The simplified expression in 2 corresponds to the expression in option B,

    The simplified expression in 3 corresponds to the expression in option A,

Therefore, the correct answer (among the options offered) is option B.

Answer

1-b, 2-c, 3-a

Exercise #7

Match the expressions that have the same value:

  1. (2x+y)(x+2y) (2x+y)(x+2y)

  2. (2x+2y)(x+y) (2x+2y)(x+y)

  3. (2x−y)(x−2y) (2x-y)(x-2y)

    a. 2x2+4xy+2y2 2x^2+4xy+2y^2

    b. 2x2−5xy+2y2 2x^2-5xy+2y^2

    c. 2x2+5xy+2y2 2x^2+5xy+2y^2

Video Solution

Step-by-Step Solution

Let's simplify the given expressions, open the parentheses using the expanded distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an inseparable part of it, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first, as an expression where addition operation exists between all terms (if needed),

We will therefore simplify each of the expressions in the given problem, while being careful about the above, first opening the parentheses using the mentioned distribution law and then using the commutative law of addition and multiplication and combining like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (2x+y)(x+2y)2x⋅x+2x⋅2y+y⋅x+y⋅2y2x2+4xy+xy+2y22x2+5xy+2y2 (2x+y)(x+2y) \\ 2x\cdot x+2x\cdot 2y+y\cdot x+y\cdot2y\\ 2x^2+4xy+xy+2y^2\\ \boxed{2x^2+5xy+2y^2}

  2. (2x+2y)(x+y)2x⋅x+2x⋅y+2y⋅x+2y⋅y2x2+2xy+2yx+2y22x2+4xy+2y2 (2x+2y)(x+y) \\ 2x\cdot x+2x\cdot y+2y\cdot x+2y\cdot y\\ 2x^2+2xy+2yx+2y^2\\ \boxed{2x^2+4xy+2y^2}

  3. (2x−y)(x−2y)↓(2x+(−y))(x+(−2y))2x⋅x+2x⋅(−2y)+(−y)⋅x+(−y)⋅(−2y)2x2−4xy−yx+2y22x2−5xy+2y2 (2x-y)(x-2y) \\ \downarrow\\ \big(2x+(-y)\big)\big(x+(-2y)\big) \\ 2x\cdot x+2x\cdot(-2y)+(-y)\cdot x+(-y)\cdot(-2y)\\ 2x^2-4xy-yx+2y^2\\ \boxed{2x^2-5xy+2y^2}

    Now, we'll use the commutative law of addition and multiplication to notice that:

    The simplified expression in 1 matches the expression in option C,

    The simplified expression in 2 matches the expression in option A,

    The simplified expression in 3 matches the expression in option B,

Therefore, the correct answer (among the suggested options) is answer A.

Answer

1-c, 2-a, 3-b

Exercise #8

Group the expressions that have the same value.

  1. (b+c)(a−4) (b+c)(a-4)

  2. (4+c)(a+b) (4+c)(a+b)

  3. (a+4)(b−c) (a+4)(b-c)

  4. (b+4)(c−a) (b+4)(c-a)

    a. ac+ab−4b−4c ac+ab-4b-4c

    b. 4b+ab−4c−ac 4b+ab-4c-ac

    c. bc−ab+4c−4a bc-ab+4c-4a

    d. 4a+4b+ac+cb 4a+4b+ac+cb

Video Solution

Answer

1-a, 2-d, 3-b, 4-c

Exercise #9

Join expressions of equal value

  1. (2a+b)(b+4) (2a+b)(b+4)

  2. (4+a)(2b+b) (4+a)(2b+b)

  3. (2a−b)(b−4) (2a-b)(b-4)

  4. (2a−b)(b+4) (2a-b)(b+4)

    a.2ab−8a−b2+4b 2ab-8a-b^2+4b

    b.12b+3ab 12b+3ab

    c.2ab+8a−b2−4b 2ab+8a-b^2-4b

    d.2ab+8a+b2+4b 2ab+8a+b^2+4b

Video Solution

Answer

1-d, 2-b, 3-a, 4-c

Exercise #10

Join expressions of equal value

  1. (2x+9)(y+4) (2x+9)(y+4)

  2. (2y+9)(x+4) (2y+9)(x+4)

  3. (2x−9)(y−4) (2x-9)(y-4)

  4. (2x+9)(y−4) (2x+9)(y-4)

    a.2xy+8x+9y+36 2xy+8x+9y+36

    b.2xy−8x+9y−36 2xy-8x+9y-36

    c.2xy−8x−9y+36 2xy-8x-9y+36

    d.2xy+8y+9x+36 2xy+8y+9x+36

Video Solution

Answer

4-b, 3-c, 2-d, 1-a

Question 1

Join expressions of equal value

  1. \( 3(y+b)+4x \)

  2. \( (3+4x)(y+b) \)

  3. \( (4y+3)(x+b) \)

    a.\( 3y+3b+4x \)

    b.\( 4yx+4yb+3x+3b \)

    c.\( 3y+3b+4xy+4xb \)

Question 2

Join expressions of equal value

  1. \( (4+x)(y+8+x) \)

  2. \( (4+x+y)(8+x) \)

  3. \( (12+x)(y+x) \)

    a.\( x^2+12x+xy+12y \)

    b.\( x^2+12x+xy+4y+32 \)

    c.\( x^2+12x+xy+8y+32 \)

Question 3

Join expressions of equal value

  1. \( (-7-x)(a-13) \)

  2. \( (-a+13)(-7-x) \)

  3. \( (7+x)(a-13) \)

    a.\( 7a+ax-91-13x \)

    b.\( -7a+91-ax+13x \)

    c.\( 7a-ax+91-13x \)

Question 4

Join expressions of equal value

  1. \( (a+g)x+3 \)

  2. \( (x+3)(a+g) \)

  3. \( (a-g)x-3 \)

    a.\( xa+xg+3a+3g \)

    b.\( ax+gx+3 \)

    c.\( ax-gx-3 \)

Question 5

Join expressions of equal value

  1. \( (m-n)(a-4) \)

  2. \( (4-n)(m+a) \)

  3. \( (n-m)(4-a) \)

    a.\( 4m+4a-nm-na \)

    b.\( ma-4m-na+4n \)

    c.\( -ma+4m+na-4n \)

Exercise #11

Join expressions of equal value

  1. 3(y+b)+4x 3(y+b)+4x

  2. (3+4x)(y+b) (3+4x)(y+b)

  3. (4y+3)(x+b) (4y+3)(x+b)

    a.3y+3b+4x 3y+3b+4x

    b.4yx+4yb+3x+3b 4yx+4yb+3x+3b

    c.3y+3b+4xy+4xb 3y+3b+4xy+4xb

Video Solution

Answer

1-a, 2-c, 3-b

Exercise #12

Join expressions of equal value

  1. (4+x)(y+8+x) (4+x)(y+8+x)

  2. (4+x+y)(8+x) (4+x+y)(8+x)

  3. (12+x)(y+x) (12+x)(y+x)

    a.x2+12x+xy+12y x^2+12x+xy+12y

    b.x2+12x+xy+4y+32 x^2+12x+xy+4y+32

    c.x2+12x+xy+8y+32 x^2+12x+xy+8y+32

Video Solution

Answer

1-c, 2-b, 3-a

Exercise #13

Join expressions of equal value

  1. (−7−x)(a−13) (-7-x)(a-13)

  2. (−a+13)(−7−x) (-a+13)(-7-x)

  3. (7+x)(a−13) (7+x)(a-13)

    a.7a+ax−91−13x 7a+ax-91-13x

    b.−7a+91−ax+13x -7a+91-ax+13x

    c.7a−ax+91−13x 7a-ax+91-13x

Video Solution

Answer

1-b, 2,3-a

Exercise #14

Join expressions of equal value

  1. (a+g)x+3 (a+g)x+3

  2. (x+3)(a+g) (x+3)(a+g)

  3. (a−g)x−3 (a-g)x-3

    a.xa+xg+3a+3g xa+xg+3a+3g

    b.ax+gx+3 ax+gx+3

    c.ax−gx−3 ax-gx-3

Video Solution

Answer

1-b, 2-a, 3-c

Exercise #15

Join expressions of equal value

  1. (m−n)(a−4) (m-n)(a-4)

  2. (4−n)(m+a) (4-n)(m+a)

  3. (n−m)(4−a) (n-m)(4-a)

    a.4m+4a−nm−na 4m+4a-nm-na

    b.ma−4m−na+4n ma-4m-na+4n

    c.−ma+4m+na−4n -ma+4m+na-4n

Video Solution

Answer

1=3=b, 2=a

Question 1

Join expressions of equal value

  1. \( (y+5)(x+7) \)

  2. \( (x+5)(y+7) \)

  3. \( (x-5)(y-7) \)

  4. \( (x-5)(y+7) \)

    a.\( xy+7y+5x+35 \)

    b.\( xy+7x+5y+35 \)

    c.\( xy-7x-5y+35 \)

    d.\( xy+7x-5y-35 \)

Exercise #16

Join expressions of equal value

  1. (y+5)(x+7) (y+5)(x+7)

  2. (x+5)(y+7) (x+5)(y+7)

  3. (x−5)(y−7) (x-5)(y-7)

  4. (x−5)(y+7) (x-5)(y+7)

    a.xy+7y+5x+35 xy+7y+5x+35

    b.xy+7x+5y+35 xy+7x+5y+35

    c.xy−7x−5y+35 xy-7x-5y+35

    d.xy+7x−5y−35 xy+7x-5y-35

Video Solution

Answer

1-a, 2-b, 3-c, 4-d