# Simultaneous Equations year 9 please solve

$$xy=4\tag1$$ $$2x - y - 7 = 0\tag2$$

Simultaneous Equations Please Solve in in year 9.

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@Hayden: Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Consider editing you question – Dennis Gulko Mar 19 '13 at 8:55

This is how I will explain to a 9 year old kid :-)

Equations

$$xy=4\tag1$$ $$y=2x-7\tag2$$

Multiplying by $x$, both sides of $(2)$

One important thing to note here is the implied assumption that $x\ne0$. If it wasn't then $y$ would be a very large number which is always greater than the number you can imagine and dividing any number with it, would be equal to $0$. Such numbers are called (Infinity) $\infty$. So from here on we will assume $x \ne 0$

$$xy = 2\cdot x^2 - 7x$$ Using Equation $(1)$ $$4 = 2\cdot x^2 - 7x$$ $$2\cdot x^2 - 7x -4=0\tag3$$ To factorize the function, express it in the form $$Ax^2+Bx+C=0$$ Find all the factors of the product $A\cdot C = 2*-4 = -8$ $${(8,-1),(-8,1),(4,2),(4,-2),(4,-2)}$$ From the above set, choose the pair $(a,b)$ that matches the equation

$$a+b=B$$

In this case it is $B=-7=-8+1$

Substitute it for B in the $(3)$ $$2\cdot x^2 + (-8+1)x -4=0$$ $$2\cdot x^2 - 8x + x -4=0$$ Group by common factors $$2x(x - 4) + 1(x -4)=0$$ $$(2x+1)(x - 4)=0$$

So, either $2x+1=0;x=-\frac{1}{2}\text{ or }x-4=0;x=4$

So, x is one of $\{-\frac{1}{2},4\}$

Now use equation (1) to determine $y$

From (1), we have $$xy=4$$ if $x = -1/2$ $$y=-8$$ if $x = 4$ $$y=1$$ 

So the solution to the given equation is one of $\{(-\frac{1}{2},-8),(4,1)\}$

A picture is worth a thousand words and here is one such picture for you

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You have numbered two of your questions as (3) -- I'm not sure whether this was on purpose. – Martin Sleziak Mar 19 '13 at 9:21
But also for a 9 year old one should mention that the solution set doesn't change after multiplication with $x$ since there is no solution for $x=0$. Otherwise there is the "famous" example where one can deduce from $x=y$ that $2=1$. – Quickbeam2k1 Mar 19 '13 at 9:23
This is what I was talking about. Very nice brother. – Inceptio Mar 19 '13 at 9:33
"Find all the factors of the product $A\cdot B = 2*4 = -8$" should be replaced with "Find all the factors of the product $A\cdot C = 2*4 = -8$". – learner Mar 19 '13 at 11:58

You are probably expected to do it like this. We have $y=2x-7$. Substitute for $y$ in $xy=4$. We get $x(2x-7)=4$. Rewrite as $2x^2-7x-4=0$.

Then use the Quadratic Formula, or factor and obtain $(2x+1)(x-4)=0$. Thus $x=-\frac{1}{2}$ or $x=4$, and now you can find the corresponding values of $y$.

Another way: Here is a different way, perhaps a little harder. Note the identity $$(2x+y)^2=(2x-y)^2+8xy.$$ From $2x-y=7$ and $xy=4$ we get that $(2x+y)^2=49+32=81$.

So $2x+y=9$ or $2x+y=-9$.

From $2x-y=7$ and $2x+y=9$, we get $x=4$, $y=1$.

From $2x-y=7$ and $2x+y=-9$ we get $x=-\frac{1}{2}$, $y=-8$.

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Using the first expression $xy = 4$, we can get an expression for $x$ in terms of $y$ i.e $x = \frac{4}{y}$. Then in the second expression substitute for the $x$. i.e the expression $2x - y -7 = 0$ would become $\frac{8}{y} - y - 7 = 0$. Further simplifying this leads to $8 - y^2 -7y = 0$ which is nothing but $y^2 + 7y - 8 = 0$. Solving this using quadratic equations or factorization method, gives the values of $x$ and $y$ as $(4 , 1)$ or $(-\frac{1}{2} , -8)$.

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you should also consider the case for $y=0$ there is no solution. Hence it is allowed to divide by $y$. – Quickbeam2k1 Mar 19 '13 at 9:17

{x = 4, y = 1} and {x = -1/2, y = -8}

For simple problems there are simple solutions: use WolframAlpha

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can you show me how you did it please kind sir? – MATHSUSER Mar 19 '13 at 8:38
@user958624: I don't think that saying what the answer is well help the OP deal with similar problems once he encounters them. An answer should be a way to solve the equations or hints to solving. The use of WA is recommended ONLY when you know how to solve it by yourself! – Dennis Gulko Mar 19 '13 at 8:58
@DennisGulko: What does he mean by year 9? Does he say that he is 9 year old? – Inceptio Mar 19 '13 at 9:03
@Inceptio: Why do you ask me? How could I know what he meant by that? :) – Dennis Gulko Mar 19 '13 at 9:07
@DennisGulko: If he meant he was 9 year old. Then, there should be a basic way of making him understand. Right? Drawing figures and all? :) – Inceptio Mar 19 '13 at 9:09