GCSEMathsEdexcelFoundationFlashcardsAlgebraSimultaneous Equations

Simultaneous Equations (Edexcel GCSE Maths: Foundation): Flashcards

Exam code: 1MA1

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FrontSimultaneous Equations
Define linear simultaneous equations.

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Cards in this collection (8)

Define linear simultaneous equations.
Linear simultaneous equations are two equations in two unknowns that are solved together, so that the solution satisfies both at the same time.
They are linear because they contain no $x^{2}$ or $y^{2}$ terms.
True or False?
There is more than one pair of values of $x$ and $y$ that satisfies $3 x + 2 y = 11$ .
True.
Both $x = 3, y = 1$ and $x = 1, y = 4$ satisfy it, and there are many more pairs that do.
This is why a second equation is needed to pin the answer down to one pair.
In $3 x + 2 y = 11$ and $2 x - y = 5$ , how do you make the $y$ terms match so that one can be eliminated?
Multiply every term of the second equation by $2$ , which gives $4 x - 2 y = 10$ .
The first equation is left alone, because its $y$ term is already $2 y$ .
Complete the rule for eliminating a term:
When the signs in front of the terms you want to eliminate are the same, $______$ the equations.
When the signs are different, $______$ the equations.
The completed rule is:
When the signs in front of the terms you want to eliminate are the same, subtract the equations.
When the signs are different, add the equations.
You have eliminated $x$ from a pair of simultaneous equations and found that $y = 1$ . How do you find $x$ ?
Substitute $y = 1$ back into one of the original equations and solve it for $x$ .
Either original equation will do, and the other can then be used to check that both values are correct.
True or False?
Subtracting $6 x - 3 y = 15$ from $6 x + 4 y = 22$ gives $y = 7$ .
False.
The $y$ terms give $4 y - (- 3 y) = 7 y$ , not $y$ , because subtracting a negative term adds it on.
The result is $7 y = 7$ , so $y = 1$ .
How do you start solving a pair of simultaneous equations by substitution instead of elimination?
Rearrange one of the equations into the form $y = \dots$ (or $x = \dots$ ), then replace every $y$ in the other equation with that expression.
Put the expression in brackets as you substitute it, so that it is treated as a single quantity.
How do you find the solution of two simultaneous equations from their graphs?
Plot both equations on the same axes and find the point where the two lines intersect.
The $x$ -coordinate of that point is the solution for $x$ , and the $y$ -coordinate is the solution for $y$ .

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