adsense

Solving SBT Lesson 1: Algebraic Expressions (C7 SBT Math 7 Horizons)

————-

### Solution 1 page 25 SBT Math 7 Creative horizon episode 2 – CTST

Write a numerical expression representing the area of a parallelogram whose base length is 6 cm and height is 5 cm.

**Detailed instructions for solving Lesson 1**

**Solution method**

Apply the formulas for the area of known planes to write the expression.

**Detailed explanation**

The area of a parallelogram is equal to the product of its base and height.

Therefore, the area of the parallelogram is \(5.6\) cm^{2}.

–>

**— *******

### Solve problem 2 page 25 SBT Math 7 Creative horizon episode 2 – CTST

Write an algebraic expression representing the number of oranges arranged in the top 4 layers of the block in Figure 1.

**Detailed instructions for solving Lesson 2**

**Solution method**

We see the number of oranges in each layer arranged in a square. So we just need to count the number of oranges on the front of the cube and use the formula to calculate the area of the square.

**Detailed explanation**

We see the number of oranges in each layer arranged in a square.

Row 1 has the number of results \({1^2}\); row 2 has the result \({2^2}\); row 3 has the number of results \({3^2}\); row 4 has a result of \({4^2}\)

The algebraic expression for the number of oranges arranged in the top 4 layers of the block is \({1^2} + {2^2} + {3^2} + {4^2}\).

–>

**— *******

### Solution 3 page 25 SBT Math 7 Creative horizon episode 2 – CTST

Write an algebraic expression for the area of a rhombus whose first diagonal is 4 cm longer than the second.

**Detailed instructions for solving Lesson 3**

**Solution method**

Step 1: Set the hide and its condition.

Step 2: Represent other quantities according to the set implicits.

Step 3: Represent the relationship between the implicit quantities by mathematical operations.

**Detailed explanation**

Let the length of the first diagonal be \(x\,\left( {x > 0} \right)\) cm.

The length of the second diagonal is \(x – 4\) cm.

The algebraic expression for the area of the rhombus is \(\frac{1}{2}.x.\left( {x – 4} \right) = \frac{1}{2}{x^2 } – 2x\) cm^{2}.

–>

**— *******

### Solution 4 page 25 SBT Math 7 Creative horizon episode 2 – CTST

Write an algebraic expression that represents the volume of a rectangular box whose length is 6 cm more than its width and 3 cm more than its height.

**Detailed instructions for solving Lesson 4**

**Solution method**

Step 1: Set the hide and its condition.

Step 2: Represent other quantities according to the set implicits.

Step 3: Represent the relationship between the implicit quantities by mathematical operations.

**Detailed explanation**

Let the length of the rectangle be \(x\,\left( {x > 0} \right)\) cm.

The width of the rectangle is \(x – 6\) cm

The height of the rectangle is \(x – 3\) cm.

The algebraic expression for the volume of a rectangular box is \(x\left( {x – 6} \right)\left( {x – 3} \right) = {x^3} – 3{x^ 2} – 6x + 18\) cm^{3}.

–>

**— *******

### Solve problem 5 pages 25 SBT Math 7 Creative horizon episode 2 – CTST

Write an algebraic expression that represents:

a) Sum of \(3{x^2} + 1\) and \(5\left( {y – 2} \right)\);

b) Sum of the squares of \(a + 2\) and \(b – 1\).

**Detailed instructions for solving Lesson 5**

**Solution method**

Express algebraic expressions by mathematical operations required by the problem.

**Detailed explanation**

a) The sum of \(3{x^2} + 1\) and \(5\left( {y – 2} \right)\) is \(\left( {3{x^2} + 1} \) right) + 5\left( {y – 2} \right)\)

adsense

b) The sum of the squares of \(a + 2\) and \(b – 1\) is \({\left( {a + 2} \right)^2} + {\left( {b – 1} \right)^2}\).

–>

**— *******

### Solution 6 page 25 SBT Math 7 Creative horizon episode 2 – CTST

The temperature in a town in the morning is x \(^\circ C\), at noon it increases by y \(^\circ C\), and in the evening it decreases by t \(^\circ C\)so with noon. Write an expression for the temperature at night. Calculate the temperature in the evening when \(x = 25;\,y = 5;\,t = 7\).

**Detailed instructions for solving lesson 6**

**Solution method**

Step 1: Represent the relationship between the quantities in the implicit by mathematical operations.

Step 2: Substitute the values of the variable to calculate the temperature in the evening.

**Detailed explanation**

The expression for the evening temperature is: \(x + y – t\).

The temperature at night when \(x = 25;\,y = 5;\,t = 7\) is \(25 + 5 – 7 = 23\)\(^\circ C\).

–>

**— *******

### Solution 7 page 25 SBT Math 7 Creative horizon episode 2 – CTST

Simplify the following expressions:

a) \(4\left( {2y – 3x} \right) – 3\left( {x – 2y} \right)\)

b) \({x^2} + 5y – 2y – 7{x^2}\)

**Detailed instructions for solving lesson 7**

**Solution method**

Use the rules of punctuation, breaking brackets learned to shorten expressions.

**Detailed explanation**

a) \(4\left( {2y – 3x} \right) – 3\left( {x – 2y} \right) = 8y – 12x – 3x + 6y = \left( {8y + 6y} \right) + \left( { – 12x – 3x} \right) = 14y – 15x\)

b) \({x^2} + 5y – 2y – 7{x^2} = \left( {{x^2} – 7{x^2}} \right) + \left( {5y – 2y} \right) = – 6{x^2} + 3y\)

–>

**— *******

### Solve problem 8 page 26 SBT Math 7 Creative horizon episode 2 – CTST

A rectangular garden has a width of a (m), length 6m more than width. The walkway is x (m) wide (see Figure 2). Write an expression that represents the area of the remainder of the garden. Calculate the area when a = 30 m, x = 1m.

**Detailed instructions for solving Lesson 8**

**Solution method**

Step 1: Set the hide and its condition.

Step 2: Represent other quantities according to the set implicits.

Step 3: Represent the relationship between the implicit quantities by mathematical operations.

**Detailed explanation**

Let the width of the rectangle be \(a\) (m) \(\left( {a > 0} \right)\)

According to the problem, we have: length is 6m more than width, so length of rectangle is \(a + 6\)(m)

The path area is \(far + x.\left( {a + 6} \right) – {x^2}\)

The area of the rectangle is \(a.\left( {a + 6} \right)\)

The remaining area is \(a.\left( {a – 6} \right) – \left[ {x.a + x.\left( {a + 6} \right) – {x^2}} \right]\).

Area left when \(a = 30\) m, \(x = 1\)m is \(a.\left( {a + 6} \right) – \left)[ {x.a + x.\left( {a + 6} \right) – {x^2}} \right] = 30.\left( {30 + 6} \right) – \left[ {1.30 + 1.\left( {30 + 6} \right) – {1^2}} \right] = 1015\) m^{2}

–>

**— *******

### Solve lesson 9 page 26 SBT Math 7 Creative horizon episode 2 – CTST

Every morning you practice walking combined with running. The walking speed is 4 km/h and the running speed is 8 km/h.

a) Write an expression representing the distance that Thu walked x hours and ran y hours.

b) Calculate the distance when \(x = 15\) minutes and \(y = 30\) minutes.

**Detailed instructions for solving Lesson 9**

**Solution method**

Step 1: Express other quantities in terms of the given unknowns.

Step 2: Represent the relationship between the quantities in the implicit by mathematical operations.

**Detailed explanation**

a) The expression that represents the distance that Thu has walked x hours and run y is now \(4.x + 8.y\).

b) Convert \(x = 15\) minutes \( = 0.25\) hours and \(y = 30\) minutes \( = 0.5\)hours.

The distance that you travel is: \(4.0.25 + 8.0.5 = 5\)km.

–>

**— *******