MHB Solve Simultaneous Equations: Find X & Y Values

AI Thread Summary
The discussion focuses on solving simultaneous equations using substitution, specifically the equations 2x + 3y = 5 and 5x - 2y = -16. The user initially struggles with the substitution method and becomes confused about combining terms and handling fractions. Clarifications are provided, explaining how to correctly combine coefficients and distribute terms, leading to the correct values of x and y, which are x = -2 and y = 3. The conversation emphasizes the importance of proper algebraic manipulation and understanding the rules for combining like terms. Ultimately, the user seeks clarity on the steps to avoid errors in future calculations.
Casio1
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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:confused:
 
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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
 
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:confused:
 
Both 5x and (4/3)x are contained in their sum of (19/3)x.

It's like if you had 2x + 3x = 5x.
 
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}
 
Re: Simultaineous Equations

Casio said:
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:confused:
 
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