Question about YouTube Videos

In the equation 6.5x + 1.4y = 59, what is the value of x when y = 5?

Ad

6.5x + 1.4y = 59

=> 6.5x + 1.4(5) = 59

=> 6.5x + 7 = 59

=> 6.5x = 59 - 7

=> 6.5x = 52

=> x = 52/6.5

=> x = 8

Posted on Dec 07, 2012

Ad

Hi there,

Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two.

Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

Here's a link to this great service

Good luck!

Posted on Jan 02, 2017

Ad

First of all you equation is not one : it has nothing on the right side of the = sign. But to answer the general question let us write the equation as **50+25x-5y=0**

**X-intercept (also know as roots) There may be several**

Definition: X-intercepts are those values of the independent variable x**for which y=0**. For a straight line there con be at most 1 x-intercept.

To find the intercept, set y=0 in the equation of the line and solve for x

50+25x-5(0)=0 or 50+25x=0. The solution is** x=-(50/25)=-2**

**Y-intercept (also know as the initial value.** There can only be 1 y-intercept, otherwise the expression does not represent a function.

Definition: It is the value of the dependent variable y when x=0 (where the function crosses the y-axis

To find it, set the x-value to 0 in the equation of the line.

**50+25x-5y=0**

50+25(0)-5y=0, or 50-5y=0. The solution is**y=50/5=10**

The straight line cuts the x-axis at the point (-2, 0) and the y-axis at the point (0,10)

Definition: X-intercepts are those values of the independent variable x

To find the intercept, set y=0 in the equation of the line and solve for x

50+25x-5(0)=0 or 50+25x=0. The solution is

Definition: It is the value of the dependent variable y when x=0 (where the function crosses the y-axis

To find it, set the x-value to 0 in the equation of the line.

50+25(0)-5y=0, or 50-5y=0. The solution is

The straight line cuts the x-axis at the point (-2, 0) and the y-axis at the point (0,10)

Jan 28, 2014 | Computers & Internet

To find the solution, first find the value of y for each equation.

Then substitue one equation into the other so that you only the x variable left.

Then just solve for x.

Once you have a value for x, then you can easily solve for y.

So for the first equation:

3y - 6x = -3

3y = 6x - 3

**y = 2x - 1**

Now for the second equation:

2y + 8x = 10

2y = -8x + 10

**y = -4x + 5**

Since both equations equal y, they also equal each other, therefore:

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1

6x = 6

**x=1**

Now substitute x=1 into either original equation:

y = 2x - 1

y = 2 (1) - 1

y = 2 - 1

**y = 1**

Therefore the solution is x=1 and y=1

Good luck, I hope that helps.

Joe.

Then substitue one equation into the other so that you only the x variable left.

Then just solve for x.

Once you have a value for x, then you can easily solve for y.

So for the first equation:

3y - 6x = -3

3y = 6x - 3

Now for the second equation:

2y + 8x = 10

2y = -8x + 10

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1

6x = 6

y = 2x - 1

y = 2 (1) - 1

y = 2 - 1

Good luck, I hope that helps.

Joe.

Nov 09, 2011 | Texas Instruments TI-84 Plus Silver...

Definition

A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income Ă· Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

1. Linear Equations y= mx + b (standard form of linear equation)

2. Quadratic Equations y= ax^2+bx+c

3. Exponential Equations y= ab^x

4. Cubic Equations y=ax^3+ bx^2+cx+d

5. Quartic Equations y= ax^4+ bx^3+ cx^2+ dx+ e

6. Equation of a circle (x-h)^2+(y-k)^2= r^2

7. Constant equation y= 9 (basically y has to equal a number for it to be a constant equation).

8. Proportional equations y=kx; y= k/x, etc.

Jun 14, 2011 | Computers & Internet

First thing to do, before you enter the value of the equation to your calculator multiply the first term((1/3x+5) from the second term(1/3x+5) to make your answer simplified.Then after that you will observed that your answer look like in this pattern AX2 +BX+C it means that the equation is in general form of quadratic equation. So, in your problem,we are looking for a value of X1 and X2,when you used your calculator in solving this problem, First On you calculator then look for the mode then press it after that look for the equation(EQN).Then from the side of the word EQN there is a corresponding number to press,from my calculator it is number 5..Then press it.After you press it, there is a equations choices,look for the equation AX2 +BX+C then press it(given number to press). Now,enter the value of A, B and C from your simplified answer. After pressing the value of A press = sign from your calculator same as B and C.After you enter all the value of A,B and C continuously press = sign from your calculator till you get the answer, the value of X1 and X2.

Jun 14, 2011 | Texas Instruments TI-30 XIIS Calculator

OK, you just have to put those values into the equation, do the math, and see if it works.

-6 * 5/6 - 7 * (-2) = 9 does it?

-5 - (-14) = -5 + 14 = 9 yes it does

You can (and probably should) always check a solution by testing this way. The definition of a solution is a value or set of values that cause the equation to work out. If the test had failed (say we had gotten -6 = 9 before, which is obviously wrong), we'd know that the values weren't a solution.

-6 * 5/6 - 7 * (-2) = 9 does it?

-5 - (-14) = -5 + 14 = 9 yes it does

You can (and probably should) always check a solution by testing this way. The definition of a solution is a value or set of values that cause the equation to work out. If the test had failed (say we had gotten -6 = 9 before, which is obviously wrong), we'd know that the values weren't a solution.

Jun 09, 2011 | Computers & Internet

You are looking for a line (y=m*x+b) and have two points. From this information you can generate two equations with two unknowns (m and b are unknown).

First plug in the first point (1,5) to the general form:

5(the y value of the point) = m*1(the x-value at this point)+b

Do the same for the second point you're given.

From here solve the first equation for m in terms of b.

Plug this value of 'm' into the second equation so you will end up with something like:

3=(something in terms of b)*(-2)+b

This final equation can be solved for b (try factoring)

You now have a value for the y-intercept. Plug that into y=m*x+b

Choose either of the two points, plug into the equation on the last line with the value of b known

You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

First plug in the first point (1,5) to the general form:

5(the y value of the point) = m*1(the x-value at this point)+b

Do the same for the second point you're given.

From here solve the first equation for m in terms of b.

Plug this value of 'm' into the second equation so you will end up with something like:

3=(something in terms of b)*(-2)+b

This final equation can be solved for b (try factoring)

You now have a value for the y-intercept. Plug that into y=m*x+b

Choose either of the two points, plug into the equation on the last line with the value of b known

You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

Apr 23, 2011 | Texas Instruments TI-83 Plus Calculator

You cannot. The TI-30XA is a simple calculator lacking any symbolic algebra or equation processing. However, if your task is "given a value for x, what is the value for y" for the equation you give, you can rearrange it a little:

Thus, if you want to determine the value of y for x = 16, key in 5 [-] 16 [÷] 4 [=] and get the result y = 1. To double check, substitute x and y in the original equation:

16 + 4 x 1 = 16 + 4 = 20.

Thus, if you want to determine the value of y for x = 16, key in 5 [-] 16 [÷] 4 [=] and get the result y = 1. To double check, substitute x and y in the original equation:

16 + 4 x 1 = 16 + 4 = 20.

Feb 12, 2011 | Texas Instruments TI-30XA Calculator

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

1st equation:

11=6+a

Everything the you do to one side of the equals sign in an equation, you have to do to the other. So let's subtract 6:

5=a

------

The second equation:

a+7=10+b

Now we know from the first equation that a= 5, so let's put that in:

5+7=10+b

Or, simplifying:

12=10+b

Let's take away 10 from both sides:

2=b

So your answer is: b=2

11=6+a

Everything the you do to one side of the equals sign in an equation, you have to do to the other. So let's subtract 6:

5=a

------

The second equation:

a+7=10+b

Now we know from the first equation that a= 5, so let's put that in:

5+7=10+b

Or, simplifying:

12=10+b

Let's take away 10 from both sides:

2=b

So your answer is: b=2

Apr 04, 2010 | Vivendi ADI English and Maths Year 5 Full...

Volume 1 x .4 + Volume 2 x .75 = 70 mL x .5

Volume 2 = 70 - Volume 1

V1 x .4 + (70-V1) x .75 = 70 mL x .5

solve for V1

Volume 2 = 70 - Volume 1

V1 x .4 + (70-V1) x .75 = 70 mL x .5

solve for V1

Sep 30, 2009 | SoftMath Algebrator - Algebra Homework...

309 people viewed this question

Usually answered in minutes!

×