Solve Simultaneous Equations: Find X & Y Values

[Casio1]

I have been trying to find the values of x and y for these simultaineous equations by the method of substitution, but I can only get so far before I become confused what to do next?

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

I worked out;

y = 1/3(5 - 2x) and combined them

y = 5x - 2/3(5 - 2x) = - 16 then I worked out the brackets

y = 5x - 10/3 + 4/3x = - 16. From here I become somewhat misunderstood as to the way things are done.

I think from here I require to convert the equations into a linear equation without the fractions, but am unsure?

(5 + 4/3)x - 10/3 = - 16. I don't know what the rule is to get to this stage from the last stage above?

If I multiply out the bracket above I get;

5x + x - 2 = - 16

6x = - 16 + 2

6x = -14

x = - 2.33

Now somewhere I am misunderstanding something because x = 2.

I also know that y = 3, but until I am clear on the method to find x there is no point me trying to go forward from finding x?

Any help much appreciated

Kind regards

Casio

 

[MarkFL]

Re: Simultaineous Equations

You were fine up to here:

(5 + 4/3)x - 10/3 = - 16

When you distributed the x, made some illegal moves...

I would combine within the parentheses to obtain:

(19/3)x - 10/3 = -16

Multiply through by 3:

19x - 10 = -48

19x = -38

x = -2

Now, use either original equation, I'll use the first, to find y:

2(-2) + 3y = 5

3y = 9

y = 3

 

[Casio1]

Re: Simultaineous Equations

Please advise just to remove some confusion on my behalf what happened to the additional x here;

5x - 10/3 + 4/3x = - 16

at this point the 5x moved to 4/3x to become;

5 + 4/3x, then that became 19/3

what rule is used to make that move?

what happened to the second x?

Kind regards

Casio

 

[MarkFL]

Both 5x and (4/3)x are contained in their sum of (19/3)x.

It's like if you had 2x + 3x = 5x.

 

[soroban]

Hello, Casio!

\text{Substitution: }\;\begin{array}{cccccc}2x + 3y &=& 5 & (1) \\ 5x - 2y &=& \text{-} 16 & (2) \end{array}

I would solve it like this . . .Solve (1) for y\!:\;y \:=\:\frac{5 - 2x}{3}

Substitute into (2): .5x - 2\left(\frac{5-2x}{3}\right) \:=\:\text{-}16

Multiply by 3: .15x - 2(5-2x) \:=\:\text{-}48

. . . . . . . . . . . .15x - 10 + 4x \:=\:\text{-}48

. . . . . . . . . . . . . . . . . . .19x \:=\:-38

. . . . . . . . . . . . . . . . . . . . x \:=\:-2Substitute into (1): .2(\text{-}2) + 3y \:=\:5

. . . . . . . . . . . . . . . . -4 + 3y \:=\:5

. . . . . . . . . . . . . . . . . . . . 3y \:=\:9

. . . . . . . . . . . . . . . . . . . . . y \:=\:3

Answer: .\begin{Bmatrix}x &=& \text{-}2 \\ y&=& 3 \end{Bmatrix}

 

[HallsofIvy]

Re: Simultaineous Equations

Please advise just to remove some confusion on my behalf what happened to the additional x here;

5x - 10/3 + 4/3x = - 16

at this point the 5x moved to 4/3x to become;

5 + 4/3

No, it didn't. It became (5+ 4/3)x

then that became 19/3

what rule is used to make that move?

Adding fractions! The coefficient of x is 5+ 4/3 and you add fractions by getting a common denominator: 5+ \frac{4}{3}= \frac{5}{1}+ \frac{4}{3}= \frac{3(5)}{3(1)}+ \frac{4}{3}= \frac{15}{3}+ \frac{4}{3}= \frac{19}{3}
(5+ \frac{4}{3})x= \frac{19}{3}x

what happened to the second x?

I don't know. Apparently, you just didn't write it!

Kind regards

Casio