Examples with solutions for Variables and Algebraic Expressions: Complete the missing numbers

Exercise #1

(a+8)(b+7)=7a+8b (a+8)(b+7)-▭=7a+8b

=? ▭=\text{?}

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Expand the expression (a+8)(b+7) (a+8)(b+7) using the distributive property.
  • Step 2: Compare the expanded terms with 7a+8b 7a + 8b to find the missing part.
  • Step 3: Identify the correct expression to fill the blank.

Now, let's work through each step:
Step 1: Expand the expression (a+8)(b+7) (a+8)(b+7) .
Using the distributive property: (a+8)(b+7)=a(b+7)+8(b+7) (a+8)(b+7) = a(b+7) + 8(b+7) =ab+7a+8b+56 = ab + 7a + 8b + 56 Step 2: Compare the expanded terms with the given equation. We have: ab+7a+8b+56=7a+8b ab + 7a + 8b + 56 - ▭ = 7a + 8b Subtract 7a+8b 7a + 8b from both sides: ab+7a+8b+567a8b=0 ab + 7a + 8b + 56 - ▭ - 7a - 8b = 0 This simplifies to: ab+56=0 ab + 56 - ▭ = 0 Step 3: Solving for the missing part, we find: =ab+56 ▭ = ab + 56 Therefore, the expression that fills in the blank is ab+56 ab+56 .

The correct choice from the options provided is:

ab+56 ab+56

Answer

ab+56 ab+56

Exercise #2

Fill in the blank:

9a+8b+7c+bc=9a+12b+7c 9a+8b+7c+_—bc=9a+12b+7c

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Compare both sides of the equation
  • Step 2: Identify and equate the coefficients of similar terms
  • Step 3: Solve for the missing term

Now, let's work through each step:
Step 1: The equation we need to balance is 9a+8b+7c+_bc=9a+12b+7c9a + 8b + 7c + \_ - bc = 9a + 12b + 7c.
Step 2: Compare terms in both expressions. We see that 9a9a and 7c7c are matched on both sides. For the bb terms, we have 8b8b on the left and 12b12b on the right, indicating that the left-hand side needs an additional 4b4b to balance bb terms.
Step 3: The missing term must cancel bcbc and add 4b4b. This means the term should be 4bc\frac{4b}{c} because 4cc=4\frac{4}{c} \cdot c = 4, thus giving us the required 4b4b.

Therefore, the solution to the problem is 4c \frac{4}{c} .

Answer

4c \frac{4}{c}

Exercise #3

+6.1x+7.4y+=4.7x0.4y △+6.1x+7.4y+☐=4.7x-0.4y

What values are possible for ☐ and △ so that the equation can be satisfied?

a. =7y \triangle=-7y
=1.4x ☐=-1.4x

b. =7.8y △=-7.8y

=1.4x ☐=-1.4x

c. =1.4x △=-1.4x

=7.8y ☐=-7.8y

d. =3.4x △=-3.4x

=7.8y+2x ☐=-7.8y+2x

Video Solution

Step-by-Step Solution

To solve this equation, we need to equate the coefficients of the terms involving xx, yy, and constants on both sides of the equation:

  • On the left-hand side (LHS), the expression is +6.1x+7.4y+\triangle + 6.1x + 7.4y + \square.
  • On the right-hand side (RHS), the expression is 4.7x0.4y4.7x - 0.4y.

1. Balancing xx coefficients:
Compare the coefficients of the terms involving xx:

  • From LHS: 6.16.1.
  • From RHS: 4.74.7.

To make the coefficients equal:

6.1x+x=4.7x6.1x + \square_x = 4.7xx=4.7x6.1x=1.4x\square_x = 4.7x - 6.1x = -1.4x.

2. Balancing yy coefficients:
Compare the coefficients of the terms involving yy:

  • From LHS: 7.47.4.
  • From RHS: 0.4-0.4.

To make the coefficients equal:

7.4y+y=0.4y7.4y + \triangle_y = -0.4yy=0.4y7.4y=7.8y\triangle_y = -0.4y - 7.4y = -7.8y.

3. Constant terms:
There are no constant terms explicitly present on either side, so no adjustment is needed for constants.

Now verify the answer choices using =7.8y\triangle = -7.8y and =1.4x\square = -1.4x:

  • Choice a: =7y \triangle=-7y , =1.4x \square=-1.4x — Not correct as =7.8y\triangle=-7.8y.
  • Choice b: =7.8y \triangle=-7.8y , =1.4x \square=-1.4x — Correct.
  • Choice c: =1.4x \triangle=-1.4x , =7.8y \square=-7.8y — Correct as they match reversed because the positions are not fixed.
  • Choice d: =3.4x \triangle=-3.4x , =7.8y+2x \square=-7.8y+2x — Correctly matches based on =3.4x\triangle=-3.4x as the sum effects result in 7.8y+2x-7.8y+2x .

The correct answer to the problem is b, c, d.

Answer

b, c, d

Exercise #4

Fill in the gap so that the equation is satisfied:

25a+34ab+23b=a+13b+13b \frac{2}{5}a+\frac{3}{4}ab+\frac{2}{3}b=_—a+\frac{1}{3}b+\frac{1}{3}b

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Compare the given equation with the incomplete side.

  • Step 2: Isolate terms involving the same variables for comparison.

  • Step 3: Calculate the required terms to balance the equation.

Now, let's work through each step:

Step 1: The equation is given:
25a+34ab+23b=_a+13b+13b \frac{2}{5}a + \frac{3}{4}ab + \frac{2}{3}b = \_ - a + \frac{1}{3}b + \frac{1}{3}b

Step 2: Combine like terms on the right side. Notice 13b+13b=23b\frac{1}{3}b + \frac{1}{3}b = \frac{2}{3}b.

The equation becomes:
25a+34ab+23b=_a+23b \frac{2}{5}a + \frac{3}{4}ab + \frac{2}{3}b = \_ - a + \frac{2}{3}b

Step 3: Compare the terms involving aa, bb, and constants:

  • Terms with aa: 25a\frac{2}{5}a should remain, meaning we need to offset the a-a.

  • To balance: a(25)=25aa(\frac{2}{5}) = \frac{2}{5}a

  • Terms with abab: 34ab\frac{3}{4}ab needs to be transferred directly.

Since - on right side requires its cancellation, a plus 34b\frac{3}{4}b on missing term substitutes division. Hence,25+34b \frac{2}{5} + \frac{3}{4}b is needed to balance and supplement.

Therefore, the solution to the problem is 25+34b \frac{2}{5} + \frac{3}{4}b , which matches choice 4.

Answer

25+34b \frac{2}{5}+\frac{3}{4}b

Exercise #5

2x(23x+47y)+67x(4+)=113x2+337x+2xy 2x(\frac{2}{3}x+\frac{4}{7}y)+\frac{6}{7}x(4+☐)=1\frac{1}{3}x^2+3\frac{3}{7}x+2xy

=? ☐=\text{?}

Video Solution

Step-by-Step Solution

To solve this equation, our goal is to determine the placeholder . Let's follow these steps:
Step 1: Expand both sides.

The left side of the equation becomes:
2x(23x+47y)+67x(4+)amp;=2x23x+2x47y+67x4+67xamp;=43x2+87xy+247x+67x. \begin{aligned} 2x\left(\frac{2}{3}x + \frac{4}{7}y\right) + \frac{6}{7}x(4+☐) &= 2x \cdot \frac{2}{3}x + 2x \cdot \frac{4}{7}y + \frac{6}{7}x \cdot 4 + \frac{6}{7}x \cdot ☐ \\ &= \frac{4}{3}x^2 + \frac{8}{7}xy + \frac{24}{7}x + \frac{6}{7}x \cdot ☐. \end{aligned}

Step 2: Simplify the right side.
The right side is already in simplified form: 113x2+337x+2xy=43x2+247x+2xy.1\frac{1}{3}x^2 + 3\frac{3}{7}x + 2xy = \frac{4}{3}x^2 + \frac{24}{7}x + 2xy.

Step 3: Equate the coefficients of like terms from both sides of the equation.

  • Compare x2x^2 coefficients: 43=43\frac{4}{3} = \frac{4}{3}

  • Compare xx coefficients: 247=247\frac{24}{7} = \frac{24}{7}

  • Compare xyxy coefficients: 87+67=2\frac{8}{7} + \frac{6}{7}☐ = 2

Step 4: Solve for the placeholder using the following xyxy equation:

87+67amp;=267amp;=28767amp;=1478767amp;=67amp;=y \begin{aligned} \frac{8}{7} + \frac{6}{7}☐ &= 2 \\ \frac{6}{7}☐ &= 2 - \frac{8}{7} \\ \frac{6}{7}☐ &= \frac{14}{7} - \frac{8}{7} \\ \frac{6}{7}☐ &= \frac{6}{7} \\ ☐ &= y \end{aligned}

Therefore, the placeholder must be filled with y \mathbf{y} for the equation to be correct.

Answer

y y

Exercise #6

y(?)+23(x+y)=23x+289y+5xy y(?)+\frac{2}{3}(x+y)=\frac{2}{3}x+2\frac{8}{9}y+5xy

Video Solution

Step-by-Step Solution

To solve this problem, we will follow these steps:

  • Step 1: Identify and expand both sides of the equation.
  • Step 2: Simplify the equation.
  • Step 3: Compare the coefficients of like terms in the equations.
  • Step 4: Solve for the missing term.

Step 1: Start by examining the equation: y(?)+23(x+y)=23x+289y+5xy y(?)+\frac{2}{3}(x+y)=\frac{2}{3}x+2\frac{8}{9}y+5xy .

Step 2: Simplify the left side of the equation:
23(x+y)=23x+23y\frac{2}{3}(x+y) = \frac{2}{3}x + \frac{2}{3}y.

Step 3: Equating both sides:
y(?)+23x+23y=23x+289y+5xyy(?) + \frac{2}{3}x + \frac{2}{3}y = \frac{2}{3}x + 2\frac{8}{9}y + 5xy.

Step 4: Compare coefficients of like terms.

  • The term xx on both sides already agrees with 23x\frac{2}{3}x.
  • The coefficient of yy on the left side is 23\frac{2}{3} and on the right side, it's 289=2692\frac{8}{9} = \frac{26}{9}.
  • The extra part in the right needs to be balanced by y(?)y(?).

Step 5: Solve for the missing term by comparing coefficients:
y(?)+23y=269y+5xyy(?) + \frac{2}{3}y = \frac{26}{9}y + 5xy.

The difference to balance the yy terms is 229y=269y23y2\frac{2}{9}y = \frac{26}{9}y - \frac{2}{3}y.

The remaining term on the right side, after matching is 5xy5xy, can be on the left as part of y(?).y(?).

Therefore, the missing term y(?)y(?) is equal to 229+5x2\frac{2}{9} + 5x.

Thus, the solution to the problem is 229+5x\boxed{2\frac{2}{9} + 5x}.

Answer

229+5x 2\frac{2}{9}+5x