Simultaneous Equation Help... Please please... solve two questions..?

Question

3) 2x + 3y = 5
5x - 2y = -16


4) 10x + 5y = 2 1/2
7x - 2y = 1/10









Please explain how do solve these plus answers


Thanks :)

thanks everyone it helped alot now i can do the rest of them thanks again... and particularly to richard and emabs ... thanks brilliant explanation

Answer 1

first you want to find the least common multiple for either your x variables or your y variables. so the LCM of 3 and 2 is 6, for example. then you multiply BOTH equations by the factor that will give you 6. so
(2x + 3y = 5)*(2) = 4x + 6y = 10
and
(5x - 2y = -16)*(3) = 15x - 6y = -48
now you can cancel the 6y and -6y by means of addition, so you are able to solve for x.
4x + 6y = 10
+ 15x - 6y = -48
= 19x =-38
then, divide both sides by 19 to get the x alone.
19/19 =1 and -38/19 = -2.
so, x = -2
after that, you plug x into one of the original equations. so..
2(-2) + 3y = 5
-4 + 3y =5
you want to get the y by itself, so add 4 and then divide by 3.
3y = 9
y = 3
so the answer is x=-2, y=3. or (-2,3)

for #4, you do the same thing, you just have fractions so you have to be extra careful. first, it is probably easier to use 5/2 instead of 2 1/2. then, find the LCM of one of your variables. the LCM of 2 and 5 is 10, thats probably the easier one. it doesn't matter which variable you eliminate first, but i generally use the variable that has the lower LCM. so your new equations are:
20x + 10y = 5
35x - 10y = 1/2
add the two equations:
55x = 11/2
divide by 55:
x = 1/10
put x back into an original equation:
10(1/10) + 5y = 5/2
1 + 5y = 5/2
get y alone:
5y = 3/2
y = 3/10
so x=1/10, y=3/10 or (1/10, 3/10)

also, you can verify these answers by putting x and y back into the equation and making sure both sides of the = are the same.
hope this helped

Answer 2

3
Multiply the first equation by 2 and the second by 3:
4x + 6y = 10
15x - 6y = -48

Add the two equations to eliminate the y terms and solve for x:
19x = -38
x = -2

Substitute the solution for x into either original equation and solve for y:
4(-2) + 6y = 10
-8 + 6y = 10
6y = 18
y = 3

4
Multiply the first equation by 2 and the second by 5:
20x + 10y = 5
35x - 10y = 1/2

Add the two equations to eliminate the y terms and solve for x:
55x = 11/2
x = 11/(2•55)
x = 1/(2•5)
x = 1/10

Substitute the solution for x into either original equation and solve for y:
20(1/10) + 10y = 5
2 + 10y = 5
10y = 3
y = 3/10

Answer 3

3) you could desire to look at the two equations to work out techniques to do away with the two x or y. in case you multiply the 1st equation by making use of two and the 2d equation by making use of 5 you've 2 new equations which once you upload them jointly make y disappear. you could then discover the value of x and from which you will discover the value of y. 2x + 3y = 5 4x + 6y = 10 5x - 2y = -sixteen 15x - 6y = -40 8 upload 4x to 15x to furnish 19x upload 6y to -6y to furnish 0 upload 10 to -40 8 to furnish -38 19x = -38 x = -2 because of the fact 2x + 3y =5 then -4 + 3y =5 3y = 9 y = 3 4) in addition, we multiply the 1st equation by making use of two and the 2d by making use of 5, and upload them jointly. 20x +10y =5 35x - 10y =a million/2 55x= 5 a million/2 x = 10 20x + 10y = 5 two hundred + 10y =5 10y = -195 y = -19 a million/2 continuously double verify your solutions with the unique equations. that's ordinary tomake a mistake with the useful and adverse values.

Answer 4

2x + 3y = 5
solve for y
3y = 5 -2x
y = (5-2x)/3

substitute (5-2x)/3 for y in the 2nd equation

5x - 2y = -16
5x - 2(5-2x)/3 = -16
make 3 the denominator for the equation on the left side
by multiply 5x by 3
[ 15x -2(5-2x)/3] = -16
15x -2(5-2x) = -16 * 3
15x- (10 - 4x) = -48
15x -10 + 4x = -48
19x -10 = -48
19x = -48 +10
19x = -38
x = -38/19
x= -2

substitute x = -2 in 1st equation to get y
2x + 3y = 5
2(-2) + 3y = 5
-4 +3y = 5
3y = 5+4
3y =9
y =9/3
y = 3

so y = 3 and x = -2

do the other one yourself to get a grip on how to solve it.
hope that helps

Answer 5

First, write your equations so the variables line up because you will be adding the equations together, eventually.
3) 2x + 3y = 5
5x - 2y = -16
Now you want to multiply each equation so 1 column will cancel itself out.
In this case, we will multiply the first equation by 2 and the second by 3.
4x + 6y = 10
15x - 6y = -48 ADD COLUMNS DOWN
-------------------------
19x + 0y = -38 or 19x = -38
x = - 2
plug this result back into either original equation and solve for y ( in this case 3)
The actual solution should be written as an 'ordered pair'
(-2,3)
Try #4 yourself {HINT :Solution is (-0.1,0.3)}

Answer 6

3/10

Answer 7

problem 3: multiply eqn. 1 by 2 and eqn. 2 by 3 and get:
4x +6y = 10
15x-6y = -48
now, add the equations and get 1 equation:
19x = - 38
x = - 2

problem 4: I don't want to do this one but I bet you could figure it out if you multiplied the first eqn. by 4 and the second eqn. by 10.