8x + 3y = 3 4x + 5y = -23

1) 2x + 5y = 7
2) 3x + 6y = 3

I'm going to use the method of elimination to solve for x and y.
Multiply 1) by 3 and 2) by 2 to allow the x's to be eliminated.

1) 6x + 15y = 21
2) 6x + 12y = 6

Now subtract line 2 from line 1.

0x + 3y = 15
---- ----
3 3 divide both sides by 3 to get y by itself.
y =5.
Substitute into 1) to calculate x.
2x + 5(5) = 7
2x + 25 = 7
2x + 25 -25 = 7 - 25
2x = -18
---- ----- divide both sides by 2 to get x by itself
2 2

x = -9

Check by plugging in answer into the other equation, in this case 2)

3 (-9) + 6(5) = 3
-27 + 30 = 3
3 = 3
We did it correctly and checked to prove that we did it right.

Good luck.

Paul

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.

Todd is 4 years old.

Let Tom's age = X
Let Todd's age = Y

From the given facts:

X = 5Y

and

X + 7 + 6 = 3 ( Y + 7 )
X+13 = 3Y + 21
X = 3Y + 8

So if:
X=5Y
and
X=3Y+8

Then:
5Y=3Y+8
2Y=8
Y=4

I hope that helps.

Joe.

Select the EQN computational mode 5:EQN

For system of linear equations in 3 unknowns select 2: in the screen below.

Arrange the variables in your equations in the same order (x first, y second, and z third).
The coefficients to enter in the editor are the factors of the variables: In your case a_1=10, b_1=-3, c_1=10, d_1=5, a_2=8, b_2=-2, c_2=9, d_2=3; a_3=8, b_3=1, c_3=-10, d_3=7.
To enter d_1, d_2, and d_3 you will have to scroll to the right to reach the cells where they should go.

Once finished entering the coefficients, press EXE to get the solutions. You may have to use the arrow Down to display the y- and z-solutions.
I verified that the matrix is non-singular and the system has a solution. In fraction form
x=27/53, y=-1 and 22/53; finally z=-23/53

Is this a maths school problem?
You need to use algebra -
2x + 5y = 232 (heena has a number of 2 and 5 coins in her purse which add up to 232)
x + y = 80 (heena has x number of 2 and y number of 5 coins which tally to 80)
therefore, x = 80 - y
so going back to first equation, 2(80 - y) + 5y = 232
160 - 2y + 5y = 232
3y = 72
y = 24
and going back to x, x = 80 - y
x = 80 - 24
x = 56
so heena has 56 rs 2, and 24 rs 5. So the answer is 24.

10x+25y

if you are looking for the answer, then here it is. Jenny gets 8 inches then Demi gets 4 inches then Drew gets 6 inches.

press y= at the top of the calculator and use the variable button2nd row 2nd button.

(13)3/8-(6)15/16