Question

# A sports store sells a total of 70- Soccer balls in one month and collects a total of $2,400. Write and Solve a System of equations to determine how many of each type of soccer ball were sold.

Hint:

### Given, Total income of the store for 70 soccer balls is $2,400.

There are two types of soccer balls and the cost of each type is different .

Frame equation considering no.of limited edition soccer balls sold be x

And no.of Pro NSL soccer ball sold be y and solve them to find x and y.

## The correct answer is: The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

### Ans :- The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

Explanation :-

Let no.of limited edition soccer balls sold be x ,no.of Pro NSL soccer ball sold be y.

Step 1:- Frame equations

Total no.of ball is 70

I.e x + y = 70 —Eq1

Total cost of balls is $2,400

Cost of x limited edition soccer balls is 65x (as per ball cost is given in diagram)

And Cost of y Pro NSL soccer ball is 15x(as per ball cost is given in diagram)

I.e 65x + 15y = 2,400 —Eq2

Step 2:- Eliminate y to find x

Do Eq2 -15(Eq1) to eliminate y

65x + 15y - 15(x+y) = 2400 - 15(70)

65x - 15x = 1350

50x = 1350 ⇒ x = 27

Step 3:- substitute value of x to find y

x + y = 70 ⇒ 27 + y = 70

⇒ y = 70 - 27

∴y = 43

∴The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

### Related Questions to study

### Ray OQ bisects ∠𝑃𝑂𝑅. Find the value of x.

- Step by step explanation:
- Given:

𝑚∠POQ = (x - 4) °

𝑚∠QOR = (5x - 20) °.

- Step 1:
- From the figure it is clear that,

∠POR is right angle hence ∠POR = 90^{o}.

and

∠POR = ∠POQ + ∠QOR

- Step 2:
- Put values of ∠COD and ∠COP

∠POR = ∠POQ + ∠QOR

90 = (x - 4) + (5x - 20)

90 = 5x + x - (20 + 4)

90 = 6x - 24

6x = 90 - 24

6x = 66

x =

x = 11

- Final Answer:

### Ray OQ bisects ∠𝑃𝑂𝑅. Find the value of x.

- Step by step explanation:
- Given:

𝑚∠POQ = (x - 4) °

𝑚∠QOR = (5x - 20) °.

- Step 1:
- From the figure it is clear that,

∠POR is right angle hence ∠POR = 90^{o}.

and

∠POR = ∠POQ + ∠QOR

- Step 2:
- Put values of ∠COD and ∠COP

∠POR = ∠POQ + ∠QOR

90 = (x - 4) + (5x - 20)

90 = 5x + x - (20 + 4)

90 = 6x - 24

6x = 90 - 24

6x = 66

x =

x = 11

- Final Answer:

### Identify all pairs of congruent angles and congruent segments.

- Step by step explanation:

- Step 1:

From the figure it is clear that

∠BAC = ∠QPR

∠BCA = ∠QRP

∠ABC = ∠PQR

This are pair of congruent angles.

- Step 2:

From the figure it is clear that

BA = QP

This is pair of congruent segments.

### Identify all pairs of congruent angles and congruent segments.

- Step by step explanation:

- Step 1:

From the figure it is clear that

∠BAC = ∠QPR

∠BCA = ∠QRP

∠ABC = ∠PQR

This are pair of congruent angles.

- Step 2:

From the figure it is clear that

BA = QP

This is pair of congruent segments.

### ∠𝐴 & ∠𝐵 are complementary. ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅ ∠𝐶

SOL – It is given that ∠𝐴 & ∠𝐵 are complementary

∠A + ∠B = 90° ---- (1)

Also, 𝐵 & ∠𝐶 are complementary

∠B + ∠C = 90° ---- (2)

From (1) and (2)

We get, ∠A + ∠B = ∠B + ∠C

∠A = ∠C

∠𝐴 ≅ ∠𝐶

Hence Proved.

### ∠𝐴 & ∠𝐵 are complementary. ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅ ∠𝐶

SOL – It is given that ∠𝐴 & ∠𝐵 are complementary

∠A + ∠B = 90° ---- (1)

Also, 𝐵 & ∠𝐶 are complementary

∠B + ∠C = 90° ---- (2)

From (1) and (2)

We get, ∠A + ∠B = ∠B + ∠C

∠A = ∠C

∠𝐴 ≅ ∠𝐶

Hence Proved.

### Given: Ray OR bisects ∠𝑃𝑂𝑆.

Prove: 𝑚∠1 = 𝑚∠2

We know that an angle bisector divides an angle into two congruent angles.

∠ POR ≅ ∠ ROS

𝑚∠1 = 𝑚∠2

Hence Proved.

### Given: Ray OR bisects ∠𝑃𝑂𝑆.

Prove: 𝑚∠1 = 𝑚∠2

We know that an angle bisector divides an angle into two congruent angles.

∠ POR ≅ ∠ ROS

𝑚∠1 = 𝑚∠2

Hence Proved.

### Solve the following by using the method of substitution

Y = - 2X-3

Y = - X-4

Explanation :-

⇒ y = -2x - 3 — eq 1

⇒ y = -x - 4—- eq 2

Step 1 :- find x by substituting y = 4x + 2 in eq 2.

-2x - 3 = -x – 4 ⇒ 2x + 3x + 4

⇒ 2x – x = 4 - 3 ⇒ x = 4 - 3

⇒ x = 1

Step 2 :- substitute value of x and find y

⇒ y = - x – 4 ⇒ y = -1 - 4

∴ y = - 5

∴ x = 1 and y = - 5 is the solution of the given pair of equations.

### Solve the following by using the method of substitution

Y = - 2X-3

Y = - X-4

Explanation :-

⇒ y = -2x - 3 — eq 1

⇒ y = -x - 4—- eq 2

Step 1 :- find x by substituting y = 4x + 2 in eq 2.

-2x - 3 = -x – 4 ⇒ 2x + 3x + 4

⇒ 2x – x = 4 - 3 ⇒ x = 4 - 3

⇒ x = 1

Step 2 :- substitute value of x and find y

⇒ y = - x – 4 ⇒ y = -1 - 4

∴ y = - 5

∴ x = 1 and y = - 5 is the solution of the given pair of equations.

### Solve the system of equations by elimination :

X - 2Y = 1

2X + 3Y= - 12

HINT: Perform any arithmetic operation and then find.

Complete step by step solution:

Let x - 2y = 1…(i)

and 2x + 3y=-12….(ii)

On multiplying (i) with 2, we get 2(x - 2y = 1)

⇒ 2x - 4y = 2…(iii)

Now, we have the coefficients of x in (ii) and (iii) to be the same.

On subtracting (ii) from (iii),

we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y

and RHS to be 2 - (- 12) = 14

On equating LHS and RHS, we have - 7y = 14

⇒ y = - 2

On substituting the value of y in (i), we get x - 2 × - 2 =1

⇒ x + 4 = 1

⇒ x = 1-4

⇒ x = - 3

Hence we get x = - 3 and y = - 2

Note: We can also solve these system of equations by making the coefficients of y

to be the same in both the equations

### Solve the system of equations by elimination :

X - 2Y = 1

2X + 3Y= - 12

HINT: Perform any arithmetic operation and then find.

Complete step by step solution:

Let x - 2y = 1…(i)

and 2x + 3y=-12….(ii)

On multiplying (i) with 2, we get 2(x - 2y = 1)

⇒ 2x - 4y = 2…(iii)

Now, we have the coefficients of x in (ii) and (iii) to be the same.

On subtracting (ii) from (iii),

we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y

and RHS to be 2 - (- 12) = 14

On equating LHS and RHS, we have - 7y = 14

⇒ y = - 2

On substituting the value of y in (i), we get x - 2 × - 2 =1

⇒ x + 4 = 1

⇒ x = 1-4

⇒ x = - 3

Hence we get x = - 3 and y = - 2

Note: We can also solve these system of equations by making the coefficients of y

to be the same in both the equations

### Solve the equation. Write a reason for each step.

𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

Opening the brackets

We get, x – 2 + 3x + 6 = 3x + 10

4x + 4 = 3x + 10 ( Adding similar terms )

4x – 3x = 10 – 4

x = 6.

### Solve the equation. Write a reason for each step.

𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

Opening the brackets

We get, x – 2 + 3x + 6 = 3x + 10

4x + 4 = 3x + 10 ( Adding similar terms )

4x – 3x = 10 – 4

x = 6.

### Use Substitution to solve each system of equations :

6X - 3Y = -6

Y = 2X + 2

Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of x in the equations .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.

Ans :- infinite no.of solutions .

Explanation :-

y = 2x + 2— eq 1

6x - 3y = -6—- eq 2

Step 1 :- find x by substituting y = 2x + 2 in eq 2.

6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6

-6 = -6

Here we get -6 = -6 which is always true i.e always having a root .

They coincide with each other and have infinite no.of solutions

They have infinite no.of solutions for the given system of equations

### Use Substitution to solve each system of equations :

6X - 3Y = -6

Y = 2X + 2

Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of x in the equations .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.

Ans :- infinite no.of solutions .

Explanation :-

y = 2x + 2— eq 1

6x - 3y = -6—- eq 2

Step 1 :- find x by substituting y = 2x + 2 in eq 2.

6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6

-6 = -6

Here we get -6 = -6 which is always true i.e always having a root .

They coincide with each other and have infinite no.of solutions

They have infinite no.of solutions for the given system of equations

### Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅

Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

𝑚∠𝑃 = 𝑚∠𝑄 ---- (1)

Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅 ---- (1)

Using transitive property which states that if A = B and B = C then A = C

We get from (1) and (2),

𝑚∠P = 𝑚∠R = 30°

Hence Proved

### Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅

Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

𝑚∠𝑃 = 𝑚∠𝑄 ---- (1)

Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅 ---- (1)

Using transitive property which states that if A = B and B = C then A = C

We get from (1) and (2),

𝑚∠P = 𝑚∠R = 30°

Hence Proved

### Use the given information and the diagram to prove the statement.

Given: 𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180° and 𝑚∠𝑃𝑀𝐶 = 150°

Prove: 𝑚∠𝑃𝑀𝐷 = 30°

SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150° ---- (1)

𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°

150° + 𝑚∠𝑃𝑀𝐷 = 180° ( - From (1) )

𝑚∠𝑃𝑀𝐷 = 180° - 150°

= 30°

Hence Proved.

### Use the given information and the diagram to prove the statement.

Given: 𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180° and 𝑚∠𝑃𝑀𝐶 = 150°

Prove: 𝑚∠𝑃𝑀𝐷 = 30°

SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150° ---- (1)

𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°

150° + 𝑚∠𝑃𝑀𝐷 = 180° ( - From (1) )

𝑚∠𝑃𝑀𝐷 = 180° - 150°

= 30°

Hence Proved.

### Solve the system of equations by elimination :

3X + 2Y = 8

X + 4Y = - 4

Let 3x + 2y = 8…(i)

and x + 4y = - 4….(ii)

On multiplying (ii) with 3, we get 3(x + 4y=-4)

⇒3x + 12y = - 12…(iii)

Now, we have the coefficients of in (i) and (iii) to be the same.

On subtracting (i) from (iii),

we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y

and RHS to be - 12 - 8 = - 20

On equating LHS and RHS, we have 10y = - 20

⇒y = - 2

On substituting the value of y in (i), we get 3x + 2× - 2 = 8

⇒ 3x - 4 = 8

⇒ 3x = 8 + 4

⇒ 3x = 12

⇒x = 4

Hence we get x = 4 and y = - 2

Note: We can also solve these system of equations by making the coefficients of y

to be the same in both the equations

### Solve the system of equations by elimination :

3X + 2Y = 8

X + 4Y = - 4

Let 3x + 2y = 8…(i)

and x + 4y = - 4….(ii)

On multiplying (ii) with 3, we get 3(x + 4y=-4)

⇒3x + 12y = - 12…(iii)

Now, we have the coefficients of in (i) and (iii) to be the same.

On subtracting (i) from (iii),

we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y

and RHS to be - 12 - 8 = - 20

On equating LHS and RHS, we have 10y = - 20

⇒y = - 2

On substituting the value of y in (i), we get 3x + 2× - 2 = 8

⇒ 3x - 4 = 8

⇒ 3x = 8 + 4

⇒ 3x = 12

⇒x = 4

Hence we get x = 4 and y = - 2

Note: We can also solve these system of equations by making the coefficients of y

to be the same in both the equations

### Give a two-column proof.

Given:

Prove: PR = 25 in

SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR

PR = PQ + QR

PR = 12 + 13

PR = 25 in.

Hence Proved.

### Give a two-column proof.

Given:

Prove: PR = 25 in

SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR

PR = PQ + QR

PR = 12 + 13

PR = 25 in.

Hence Proved.

### Use Substitution to solve each system of equations :

- 3X - Y = 7

X + 2Y = 6

Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of x in the equations.

Ans :- x = -4; y = 5

Explanation :-

- 3x - y = 7 ⇒ - 3x - 7 = y — eq 1

x + 2y = 6 —- eq 2

x + 2 ( - 3x - 7) = 6 ⇒ x - 6x - 14 = 6

⇒ - 5x = 14 + 6 ⇒ - 5x = 20

⇒ x = -4

Step 2 :- substitute value of x and find y

⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7

⇒ y = 12 - 7

∴ y = 5

x = -4 and y = 5 is the solution of the given pair of equations.

### Use Substitution to solve each system of equations :

- 3X - Y = 7

X + 2Y = 6

Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of x in the equations.

Ans :- x = -4; y = 5

Explanation :-

- 3x - y = 7 ⇒ - 3x - 7 = y — eq 1

x + 2y = 6 —- eq 2

x + 2 ( - 3x - 7) = 6 ⇒ x - 6x - 14 = 6

⇒ - 5x = 14 + 6 ⇒ - 5x = 20

⇒ x = -4

Step 2 :- substitute value of x and find y

⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7

⇒ y = 12 - 7

∴ y = 5

x = -4 and y = 5 is the solution of the given pair of equations.

### If Q is the midpoint of PR, prove that PR = 2 PQ.

Give a two-column proof.

SOL – If Q is mid – point Þ PQ = QR ---- (1)

Using Segment Addition Postulate,

We get, PR = PQ + QR

PR = PQ + PQ

PR = 2 PQ

Hence Proved.

### If Q is the midpoint of PR, prove that PR = 2 PQ.

Give a two-column proof.

SOL – If Q is mid – point Þ PQ = QR ---- (1)

Using Segment Addition Postulate,

We get, PR = PQ + QR

PR = PQ + PQ

PR = 2 PQ

Hence Proved.

### Compare 𝑚∠𝑃𝑂𝑄𝑎𝑛𝑑∠𝑆𝑂𝑇.

- Step by step explanation:
- Step 1:
- Find ∠POQ:

- Align the protractor with the ray OP on 0
^{o}as shown above. - Start reading the inner scale from the 0°.
- Step 2:
- From the figure we can see that ray OP is aligned on mark 0
^{o}. And ray OQ is aligned on mark 30^{o}.

^{o}- 0

^{o}).

∠POQ = 30

^{o}.

- Step 3:
- Find ∠SOT:
- Align the protractor with the ray OT on 0
^{o}as shown above. - Start reading the outer scale from the 0°.
- Step 4:
- From the figure we can see that ray OT is aligned on mark 0
^{o}. And the ray OS is aligned on mark 20^{o}.

- Align the protractor with the ray OT on 0

^{o}.

∠SOT = 20^{o}.

- Step 5:
- Compare ∠POQ and ∠SOT

^{o}.

- Final Answer:

^{o}.