How many more games will Aditi have to win to have an 80% winning record?

Question

The linear equation that models this is___________

Raj and Aditi like to play chess against each other. Aditi has won 12 of the 20 games so far. They decide to play more games.

How many more games will Aditi have to win to have a 70% winning record?

12 + x = 0.70(20 + x)
x = x equals StartFraction 20 Over 3 EndFraction is approximately 6.67. ≈ 6.67
She will have to win 7 more games to have a 70% winning record.

Aditi would have to win
20
more games to have an 80% winning record.

Answer 1

To determine how many more games Aditi needs to win to achieve an 80% winning record, let's define:

• Let $x$ be the number of additional games Aditi wins.
• Currently, Aditi has won 12 games and played a total of 20 games.

After winning $x$ more games, her total number of wins will be $12+x$ and her total number of games played will be $20+x$.

We want Aditi's winning percentage to be 80%, which can be set up in the following equation:

$$\frac{12+x}{20+x}=0.80$$

Now, we solve for $x$:

1. Multiply both sides by $20+x$ to eliminate the fraction: $$12+x=0.80(20+x)$$

2. Distribute $0.80$: $$12+x=16+0.80x$$

3. Rearranging the equation gives: $$12+x-0.80x=16$$

4. Combine like terms: $$12+0.20x=16$$

5. Subtract 12 from both sides: $$0.20x=4$$

6. Divide both sides by 0.20: $$x=\frac{4}{0.20}=20$$

So, Aditi would need to win 20 more games to have an 80% winning record.

The linear equation that models this situation is:

$$12+x=0.80(20+x)$$

Answer 2

solve the equation
6.8x+9.3=-2.6-17x

Answer 3

To solve the equation $6.8x+9.3=-2.6-17x$, we will follow these steps:

1. Move all terms involving $x$ to one side of the equation, and constant terms to the other side.

Starting with the original equation: $$6.8x+9.3=-2.6-17x$$

2. Add $17x$ to both sides to get all $x$ terms on the left side: $$6.8x+17x+9.3=-2.6$$

3. Combine the $x$ terms: $$(6.8+17)x+9.3=-2.6$$ $$23.8x+9.3=-2.6$$

4. Now, subtract $9.3$ from both sides: $$23.8x=-2.6-9.3$$

5. Combine the constants on the right side: $$23.8x=-11.9$$

6. Finally, divide both sides by $23.8$: $$x=\frac{-11.9}{23.8}$$

Calculating that: $$x\approx -0.499$$

So, the solution to the equation $6.8x+9.3=-2.6-17x$ is approximately: $$x\approx -0.499$$

Answer 4

distribute
combine terms
add 17x
subtract 9.3
divide 23.8
and get -0.5!! the answer

Heidi solved the equation
3(x + 4) + 2 = 2 + 5(x – 4). Her steps are below:

3x + 12 + 2 = 2 + 5x – 20
3x + 14 = 5x – 18
14 = 2x – 18
32 = 2x
16 = x
Use the drops-downs to justify how Heidi arrived at each step.

Step 1:
distributive property

Step 2:
combine like terms

Step 3:

Step 4:

Step 5:

Answer 5

Let's analyze Heidi's steps and fill in the justifications for each. Here's how we can justify each part of her solution:

Step 1:

Heidi starts with: $$3(x+4)+2=2+5(x-4)$$ Justification: Distributive property - She distributes $3$ over $(x+4)$ and $5$ over $(x-4)$.

Step 2:

She simplifies the equation: $$3x+12+2=2+5x-20$$ to: $$3x+14=5x-18$$ Justification: Combine like terms - She combined $12$ and $2$ on the left side, and $2$ and $-20$ on the right side.

Step 3:

From: $$3x+14=5x-18$$ to: $$14=5x-3x-18$$ Justification: Subtraction property of equality - She subtracts $3x$ from both sides to isolate terms involving $x$ on one side.

Step 4:

Rearranging the equation gives: $$14+18=2x$$ Justification: Addition property of equality - She adds $18$ to both sides to combine constants.

Step 5:

Finally, she computes: $$32=2x$$ and then: $$x=16$$ Justification: Division property of equality - She divides both sides by $2$ to solve for $x$.

Here’s a summary of the justifications:

• Step 1: Distributive property
• Step 2: Combine like terms
• Step 3: Subtraction property of equality
• Step 4: Addition property of equality
• Step 5: Division property of equality

Let me know if you need any more information or assistance!

Answer 6

What is the value of n in the equation –negative StartFraction one-half EndFraction left-parenthesis 2 n plus 4 right-parenthesis plus 6 equals negative 9 plus 4 left parenthesis 2 n plus 1 right-parenthesis.(2n + 4) + 6 = –9 + 4(2n + 1)?

n =

Answer 7

To solve the equation

$$-\frac{1}{2}(2n+4)+6=-9+4(2n+1),$$

let’s break it down step by step.

1. Distribute $-\frac{1}{2}$ on the left side: $$-\frac{1}{2}(2n+4)=-\frac{1}{2}\cdot 2n-\frac{1}{2}\cdot 4=-n-2.$$ Therefore, the left side becomes: $$-n-2+6=-n+4.$$

2. Distribute $4$ on the right side: $$4(2n+1)=4\cdot 2n+4\cdot 1=8n+4.$$ Therefore, the right side becomes: $$-9+8n+4=8n-5.$$

Now we re-write the equation: $$-n+4=8n-5.$$

3. Add $n$ to both sides: $$4=9n-5.$$

4. Add $5$ to both sides: $$4+5=9n,$$ which simplifies to: $$9=9n.$$

5. Divide by $9$ to solve for $n$: $$n=1.$$

So the value of $n$ is

$$\boxed{{\displaystyle 1}}.$$

Answer 8

Yes the answer is 1 also i liked any of the mini bot comments if they CORRECT

Which of these equations have no solution? Check all that apply.

2(x + 2) + 2 = 2(x + 3) + 1
2x + 3(x + 5) = 5(x – 3)
4(x + 3) = x + 12
4 – (2x + 5) = (–4x – 2)
5(x + 4) – x = 4(x + 5) – 1

Answer 9

To determine which of these equations have no solution, we need to solve each equation and analyze the results. An equation has no solution if it simplifies to a false statement (like $0=1$).

1. Equation:

$$2(x+2)+2=2(x+3)+1$$ Expanding: $$2x+4+2=2x+6+1⟹2x+6=2x+7$$ Subtract $2x$ from both sides: $$6=7(\text{False})$$ This equation has no solution.

2. Equation:

$$2x+3(x+5)=5(x-3)$$ Expanding: $$2x+3x+15=5x-15⟹5x+15=5x-15$$ Subtract $5x$ from both sides: $$15=-15(\text{False})$$ This equation has no solution.

3. Equation:

$$4(x+3)=x+12$$ Expanding: $$4x+12=x+12$$ Subtract $x$ from both sides: $$3x+12=12$$ Subtract $12$ from both sides: $$3x=0⟹x=0$$ This equation has a solution ($x=0$).

4. Equation:

$$4-(2x+5)=(-4x-2)$$ Expanding: $$4-2x-5=-4x-2⟹-2x-1=-4x-2$$ Add $4x$ to both sides: $$2x-1=-2$$ Add $1$ to both sides: $$2x=-1⟹x=-\frac{1}{2}$$ This equation has a solution ($x=-\frac{1}{2}$).

5. Equation:

$$5(x+4)-x=4(x+5)-1$$ Expanding: $$5x+20-x=4x+20-1⟹4x+20=4x+19$$ Subtract $4x$ from both sides: $$20=19(\text{False})$$ This equation has no solution.

Summary of Results:

The equations that have no solutions are:

• $2(x+2)+2=2(x+3)+1$
• $2x+3(x+5)=5(x-3)$
• $5(x+4)-x=4(x+5)-1$

Thus, the equations with no solutions are:

• 1, 2, and 5.

Answer 10

Which equation is an identity?

3(x – 1) = x + 2(x + 1) + 1
x – 4(x + 1) = –3(x + 1) + 1
2x + 3 = 2 x plus 3 equals StartFraction one-half EndFraction left-parenthesis 4 x plus 2 right-parenthesis plus 2.(4x + 2) + 2
StartFraction one-third EndFraction left-parenthesis 6 x minus 3 right-parenthesis equals 3 left-parenthesis x plus 1 right-parenthesis minus x minus 2.(6x – 3) = 3(x + 1) – x – 2

Answer 11

To determine which equation is an identity, we will analyze each equation to see if both sides are equivalent for all values of $x$. An equation is considered an identity if it simplifies to a true statement (such as $0=0$) for all $x$.

1. Equation:

$$3(x-1)=x+2(x+1)+1$$ Expanding: Left side: $$3x-3$$ Right side: $$x+2x+2+1=3x+3$$ Setting the simplified forms equal: $$3x-3=3x+3$$ Subtract $3x$ from both sides: $$-3=3(\text{False})$$ This equation is not an identity.

2. Equation:

$$x-4(x+1)=-3(x+1)+1$$ Expanding: Left side: $$x-4x-4=-3x-4$$ Right side: $$-3x-3+1=-3x-2$$ Setting the simplified forms equal: $$-3x-4=-3x-2$$ Subtract $-3x$ from both sides: $$-4=-2(\text{False})$$ This equation is not an identity.

3. Equation:

$$2x+3=\frac{1}{2}(4x+2)+2$$ Expanding: Left side: $$2x+3$$ Right side: $$\frac{1}{2}\cdot 4x+\frac{1}{2}\cdot 2+2=2x+1+2=2x+3$$ Setting the simplified forms equal: $$2x+3=2x+3$$ This holds true for all $x$, thus this equation is an identity.

4. Equation:

$$\frac{1}{3}(6x-3)=3(x+1)-x-2$$ Expanding: Left side: $$\frac{1}{3}\cdot 6x-\frac{1}{3}\cdot 3=2x-1$$ Right side: $$3x+3-x-2=2x+1$$ Setting the simplified forms equal: $$2x-1=2x+1$$ Subtract $2x$ from both sides: $$-1=1(\text{False})$$ This equation is not an identity.

Summary:

The equation that is an identity is: $$2x+3=\frac{1}{2}(4x+2)+2$$

Thus, the answer is the third equation.