Define simultaneous equations .

Simultaneous equations are two or more equations that must be satisfied by the same values at the same time. With two unknowns you need two equations, and solving them means finding the pair of values making both true.

Which equation do you substitute into which?

Rearrange the linear equation and substitute it into the quadratic one. Going the other way does not work, because putting a quadratic expression into the linear equation leaves you with something no simpler than you started with.

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Simultaneous Equations (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontLinear Simultaneous Equations
Define simultaneous equations.

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Define simultaneous equations.
Simultaneous equations are two or more equations that must be satisfied by the same values at the same time.
With two unknowns you need two equations, and solving them means finding the pair of values making both true.
What is the first step of the elimination method?
Multiply one or both equations by a constant so that the coefficients of one unknown match.
For $3 x + 2 y = 11$ and $2 x - y = 5$ , multiplying by 2 and 3 turns both $x$ terms into $6 x$ .
True or False?
To eliminate a variable whose coefficients are $+ 6$ and $- 6$ , you subtract one equation from the other.
False.
When the matching coefficients have different signs you add the equations, since $6 x + (- 6 x) = 0$ .
You subtract only when the signs are the same.
Elimination has given $y = 1$ . Complete the rest of the solution for $3 x + 2 y = 11$ :
$3 x + 2 (______) = 11$ , so $x =______$ .
Finish by checking both values in the other original equation.
The completed working is:
$3 x + 2 (1) = 11$ , so $x = 3$ .
Checking in $2 x - y = 5$ gives $2 (3) - 1 = 5$ , which confirms both values are right.
How does the substitution method work?
Rearrange one equation to make an unknown the subject, such as $y = 2 x - 5$ .
Put that expression into the other equation, which leaves a single equation in one unknown, then work back to find the second unknown.
You rearrange $2 x - y = 5$ into $y = 2 x - 5$ . Which equation do you substitute this into?
The other equation, $3 x + 2 y = 11$ .
Putting it back into the equation it came from just returns a statement that is always true, which tells you nothing about $x$ .
Why substitute $y = 2 x - 5$ into $3 x + 2 y = 11$ as $3 x + 2 (2 x - 5)$ , with brackets?
The brackets make the 2 multiply both terms, giving $4 x - 10$ rather than just $4 x$ .
Without them the constant is left behind and the equation is no longer equivalent to the one you started with.
What do the solutions of a pair of linear simultaneous equations mean on a graph?
They are the coordinates of the point where the two lines cross.
So solving $2 x - y = 3$ and $3 x + y = 4$ gives the intersection $(2, 1)$ , meaning $x = 2$ and $y = 1$ .
What makes a pair of simultaneous equations quadratic?
One of them contains an $x^{2}$ , a $y^{2}$ or an $x y$ term, so it is not a straight line.
They are also called non-linear simultaneous equations.
Which equation do you substitute into which?
Rearrange the linear equation and substitute it into the quadratic one.
Going the other way does not work, because putting a quadratic expression into the linear equation leaves you with something no simpler than you started with.
True or False?
$(2 y + 6)^{2} = 4 y^{2} + 36$
False.
A squared bracket must be expanded as a double bracket, giving $4 y^{2} + 24 y + 36$ .
Squaring the terms one at a time loses the middle term completely, which changes the equation you go on to solve.
Complete the substitution for $x^{2} + y^{2} = 36$ and $x = 2 y + 6$ :
$(______)^{2} + y^{2} = 36$
The completed substitution is:
$(2 y + 6)^{2} + y^{2} = 36$
The linear equation is already in the form $x = \dots$ , so it replaces $x$ directly, in brackets.
True or False?
A pair of quadratic simultaneous equations can have two completely different solution pairs.
True.
A line can cut a curve at two points, and each crossing gives its own pair of values.
For example $x^{2} + y^{2} = 25$ with $y - 2 x = 5$ gives both $x = 0$ , $y = 5$ and $x = - 4$ , $y = - 3$ .
Why must you keep track of which $x$ value belongs with which $y$ value?
Because each pair is one point of intersection, and only those combinations satisfy both equations.
Mixing values from different pairs gives a point that lies on neither graph, even though both numbers came from correct working.
You have solved the quadratic and found the values of $y$ . How do you find the matching values of $x$ ?
Substitute each $y$ back into the linear equation, which is much less work than using the quadratic one.
Then check each finished pair in both original equations.

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