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4 years ago

7

How do you solve

ac{x}{3} + y = 1" alt=" \frac{x}{3} + y = 1" align="absmiddle" class="latex-formula"> ?? Please help!

1 answer:

Nonamiya [84]4 years ago

6 0

Solve the following system:
{x + 3 y = 8
{4 x - y = 6
Solve for x:
x/3 + y = 1
Subtract y from both sides:
x/3 = 1 - y
Multiply both sides by 3:
Answer: x = 3 - 3 y

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3 years ago

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4 years ago

Anmol got 60 marks out of 80 marks in mathematics what percent marks did he get

Kay [80]

Answer:

75%

Step-by-Step Explanation:

There are two ways to go about this.

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Then since a fraction sign also means division, we divide 60 by 80 = 0.75

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So our answer is 75%

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3 years ago

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Artyom0805 [142]

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Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

358 tickets to the school basketball game on Friday were sold.Students tickets were $1.50 and non-student tickets were $3.25. Th

GuDViN [60]

235 student tickets and 123 non-student tickets were sold.

Step-by-step explanation:

Given,

Total number of tickets sold = 358

Total revenue generated = $752.25

Cost of student ticket = $1.50

Cost of non-student ticket = $3.25

Let,

x represent the number of student tickets sold.

y represent the number of non-student tickets sold.

According to given statement;

x+y=358 Eqn 1

1.50x+3.25y=752.25 Eqn 2

Multiplying Eqn 1 by 1.50

$1.50(x+y=358)\\1.50x+1.50y=537\ \ \ Eqn 3\\$

Subtracting Eqn 3 from Eqn 2

$(1.50x+3.25y)-(1.50x+1.50y)=752.25-537\\1.50x+3.25y-1.50x-1.50y=215.25\\1.75y=215.25$

Dividing both sides by 1.75

$\frac{1.75y}{1.75}=\frac{215.25}{1.75}\\y=123$

Putting y=123 in Eqn 1

$x+123=358\\x=358-123\\x=235$

235 student tickets and 123 non-student tickets were sold.

Keywords: linear equation, elimination

Learn more about elimination method at:

#LearnwithBrainly

6 0

3 years ago