True or False? One equation is enough to find two unknowns.

False. A single equation in two unknowns has infinitely many solution pairs. A second equation normally narrows those down to just one pair, which is why they are solved together.

A stall sells 6 bangers and 12 sparklers for £9, and 9 bangers and 10 sparklers for £12.30. To find the cost of 5 bangers and 15 sparklers, why must you first find the cost of one of each?

Because there is no simple way to scale 6 or 9 up to 5, or 12 or 10 up to 15, so the given totals cannot be combined directly. Finding the price of a single banger and a single sparkler lets you cost any combination you like.

How can you be sure your simultaneous-equation solutions are right?

Substitute both values into both original equations and check that each one balances. Checking only the equation you used to find the second value proves nothing, since that one was bound to work.

When solving a linear and a quadratic equation together, what does a repeated root mean?

The line is a tangent to the curve, touching it at exactly one point. If instead the quadratic has no roots, the line and the curve do not meet at all.

How many equations do you need to find three unknowns?

You need three equations, one for each unknown. The methods are the same as for two unknowns, either elimination or substitution.

True or False? It does not matter which of the three variables you eliminate first.

True. Any of them will work, so choose whichever is easiest to remove. A variable that already has a coefficient of 1, or that appears with opposite signs, saves the most work.

Why does combining an equation with itself get you nowhere?

Because the result is just a multiple of the same equation, so it carries no new information. Each new equation you form has to come from a different pair, which is why keeping them clearly numbered matters.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsSimultaneous Equations

Simultaneous Equations (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontLinear Simultaneous Equations
True or False?
One equation is enough to find two unknowns.

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Cards in this collection (20)

True or False?
One equation is enough to find two unknowns.
False.
A single equation in two unknowns has infinitely many solution pairs.
A second equation normally narrows those down to just one pair, which is why they are solved together.
How do you eliminate a variable from $3 x + 2 y = 11$ and $2 x - y = 5$ ?
Multiply the second equation by 2 to get $4 x - 2 y = 10$ , so that both have a $2 y$ term.
Adding the two equations then gives $7 x = 21$ , so $x = 3$ .
Complete the rule for combining the two equations:
Add them if the matching terms have $______$ signs, and subtract if they have $______$ signs.
The completed rule is:
Add them if the matching terms have opposite signs, and subtract if they have the same sign.
Either way, the aim is to make that variable disappear.
How do you solve $3 x + 2 y = 11$ and $2 x - y = 5$ by substitution?
Rearrange one equation to $y = 2 x - 5$ , then replace $y$ in the other, keeping it in brackets: $3 x + 2 (2 x - 5) = 11$ .
Solving that gives $x = 3$ , and substituting back gives $y = 1$ .
A stall sells 6 bangers and 12 sparklers for £9, and 9 bangers and 10 sparklers for £12.30. To find the cost of 5 bangers and 15 sparklers, why must you first find the cost of one of each?
Because there is no simple way to scale 6 or 9 up to 5, or 12 or 10 up to 15, so the given totals cannot be combined directly.
Finding the price of a single banger and a single sparkler lets you cost any combination you like.
Six bangers and twelve sparklers cost £9; nine bangers and ten sparklers cost £12.30. Find the cost of five bangers and fifteen sparklers.
Solving $6 B + 12 S = 9$ and $9 B + 10 S = 12.3$ gives $B = 1.2$ and $S = 0.15$ .
Then $5 \times 1.2 + 15 \times 0.15 = 8.25$ , so five bangers and fifteen sparklers cost £8.25.
How can you be sure your simultaneous-equation solutions are right?
Substitute both values into both original equations and check that each one balances.
Checking only the equation you used to find the second value proves nothing, since that one was bound to work.
What makes a pair of simultaneous equations quadratic?
One of them contains an $x^{2}$ , a $y^{2}$ or an $x y$ term.
They are also called non-linear simultaneous equations.
Complete the rule for quadratic simultaneous equations:
Substitute the $______$ equation into the $______$ equation.
The completed rule is:
Substitute the linear equation into the quadratic equation, never the other way round.
Rearrange the linear one to $y = \dots$ first, then replace every $y$ , keeping it in brackets.
Solve $x^{2} + y^{2} = 25$ and $y - 2 x = 5$ .
Substituting $y = 2 x + 5$ gives $x^{2} + {(2 x + 5)}^{2} = 25$ , which simplifies to $5 x^{2} + 20 x = 0$ .
That gives $x = 0$ or $x = - 4$ , so the solutions are $x = 0, y = 5$ and $x = - 4, y = - 3$ .
True or False?
$x^{2} + y^{2} = 25$ can be rewritten as $x + y = 5$ .
False.
You cannot square-root a sum term by term, so $\sqrt{x^{2} + y^{2}}$ is not $x + y$ .
Test it with $x = 3$ and $y = 4$ : the first equation holds, but $3 + 4 = 7$ , not 5.
Why must you say which $x$ goes with which $y$ ?
Because each solution is a point where the line meets the curve, so the two values belong together.
Listing all the $x$ values and all the $y$ values separately does not say which point is which.
When solving a linear and a quadratic equation together, what does a repeated root mean?
The line is a tangent to the curve, touching it at exactly one point.
If instead the quadratic has no roots, the line and the curve do not meet at all.
How do you solve a linear and a quadratic equation from their graphs?
Plot both on the same axes and find where they intersect.
The coordinates of each intersection are a solution pair, so $y = x^{2} + 3 x + 1$ and $y = 2 x + 1$ meeting at $(- 1, - 1)$ and $(0, 1)$ give both solutions.
How many equations do you need to find three unknowns?
You need three equations, one for each unknown.
The methods are the same as for two unknowns, either elimination or substitution.
Complete the strategy for three simultaneous equations:
Eliminate the same variable from two different pairs, leaving $______$ equations in $______$ unknowns.
The completed strategy is:
Eliminate the same variable from two different pairs, leaving two equations in two unknowns.
Solve those as usual, then substitute back into any original equation to find the third unknown.
True or False?
It does not matter which of the three variables you eliminate first.
True.
Any of them will work, so choose whichever is easiest to remove.
A variable that already has a coefficient of 1, or that appears with opposite signs, saves the most work.
Solve $2 x + 3 y - z = 17$ , $x - 3 y + 2 z = - 12$ and $3 x + y + z = 9$ .
Adding the first and third removes $z$ , giving $5 x + 4 y = 26$ , and the second plus twice the first gives $5 x + 3 y = 22$ .
Solving those gives $x = 2$ and $y = 4$ , and substituting back gives $z = - 1$ .
How does the substitution method work with three unknowns?
Rearrange one equation to make one variable the subject, such as $z = 2 x + 3 y - 17$ .
Substitute that into both of the other two equations, and the $z$ disappears from each.
Why does combining an equation with itself get you nowhere?
Because the result is just a multiple of the same equation, so it carries no new information.
Each new equation you form has to come from a different pair, which is why keeping them clearly numbered matters.

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