Simultaneous Equations(GCSE Maths)

Because the equations have $+4y$ and $-4y$, you can add the two equations together to eliminate the $y$ terms.

Divide both sides by 6.

Substitute $x = 3$ into the first equation (or the second) to solve for $y$.

Because both equations contain exactly $-4y$, you can subtract the second equation from the first to eliminate the $y$ terms.

Divide both sides by 3.

Substitute $x = 7$ into the second equation (or the first) to solve for $y$.

Because both equations have a single positive $x$, you can subtract the second equation from the first to eliminate the $x$ terms.

Divide both sides by 4.

Substitute $y = 9$ into the second equation (or the first) to solve for $x$.

Because the equations have $+y$ and $-y$, you can add the two equations together to eliminate the $y$ terms.

Divide both sides by $8$.

Substitute $x = 2.5$ into the first equation (or the second) to solve for $y$.

Fractions $x = \frac{5}{2}$ and $y = -\frac{3}{2}$ are also acceptable.

Because both equations contain exactly $+3y$, you can subtract the second equation from the first to eliminate the $y$ terms.

Divide both sides by $3$.

Substitute $x = 7$ into the second equation (or the first) to solve for $y$.

Replace $y$ in the first equation with the expression $(2x - 3)$:

Expand the bracket:

Combine the $x^2$ terms and bring $41$ to the left side:

• Factorise the quadratic equation.
• Look for two numbers that multiply to $(5 \times -32 = -160)$ and add to $-12$.
• These numbers are $-20$ and $8$.

This gives the solutions for $x$:

Substitute the $x$ values back into the linear equation ($y = 2x - 3$):

• When $x = 4$: $y = 2(4) - 3 = 8 - 3 = 5$
• When $x = -1.6$: $y = 2(-1.6) - 3 = -3.2 - 3 = -6.2$

Rearrange the linear equation to make $y$ the subject:

Substitute this expression for $y$ into the quadratic equation:

Expand and simplify to form a quadratic equation equal to zero:

Move everything to one side to make the $x^2$ term positive:

Factorise the quadratic (or use the quadratic formula):

Therefore, $x = \frac{19}{3}$ or $x = 3$.

Substitute the $x$ values back into the rearranged linear equation ($y = 7 - 2x$):

• When $x = 3$: $y = 7 - 2(3) = 7 - 6 = 1$
• When $x = \frac{19}{3}$: $y = 7 - 2\left(\frac{19}{3}\right) = \frac{21}{3} - \frac{38}{3} = -\frac{17}{3}$

Subtract $2x$ and add $1$ to both sides:

Factorise the quadratic equation:

So, $x = 3$ or $x = 1$.

Substitute the $x$ values back into the linear equation ($y = 2x - 1$):

• When $x = 3$: $y = 2(3) - 1 = 6 - 1 = 5$
• When $x = 1$: $y = 2(1) - 1 = 2 - 1 = 1$

Subtract $x$ and add $5$ to both sides:

Rearrange to $x^2 = 9$ or factorise as $(x - 3)(x + 3) = 0$:

Substitute the $x$ values back into the linear equation ($y = x - 5$):

• When $x = 3$, $y = 3 - 5 = -2$
• When $x = -3$, $y = -3 - 5 = -8$

Since both equations equal $y$, set them equal to each other:

Subtract $x$ and subtract $3$ from both sides:

Factorize the quadratic equation:

So, $x = -5$ or $x = 1$.

Substitute the $x$ values back into the linear equation ($y = x + 3$):

• When $x = -5$: $y = -5 + 3 = -2$
• When $x = 1$: $y = 1 + 3 = 4$

Let $a$ be the length of route A and $b$ be the length of route B.

• From the information that route B is $8\text{ miles}$ longer than route A, we have: $b = a + 8$.
• From the total distance jogged ($9$ times route A and $5$ times route B equals $89\text{ miles}$), we have: $9a + 5b = 89$.

Substitute $b = a + 8$ into the second equation:

Route C is $20\%$ longer than route B, so we calculate $120\%$ of route B's length:

Let $a$ be the cost of an adult ticket and $c$ be the cost of a child ticket.

• Equation 1 (Fozia):

• Equation 2 (Sami):

Multiply Equation 1 by 3 so that the coefficient of $c$ matches Equation 2:

Subtract the original Equation 2 from the new Equation 1:

Substitute $a = 27$ into the original Equation 1:

Let $s$ represent the number of sweets in a small pack, and $b$ represent the number of sweets in a big pack. This gives us Equation 1:

And Equation 2:

To eliminate $b$, multiply Equation 1 by $2$, and Equation 2 by $3$:

Subtract the new Equation 1 from the new Equation 2 to eliminate $b$:

Substitute $s = 16$ into the original Equation 1:

Make $y$ the subject to substitute it into the quadratic equation later (making $x$ the subject also works):

Replace $y$ in the first equation with your new expression:

Square the fraction:

Multiply the entire equation by $4$ to clear the denominator:

Expand the bracket, being careful with the negative sign:

Rearrange to form a standard quadratic equation equal to zero (moving terms to the right side makes $x^2$ positive):

Factorise the quadratic equation by finding two numbers that multiply to $65$ and add to $-18$ ($-5$ and $-13$):

Therefore, the solutions for $x$ are:

Substitute $x$ back into the rearranged linear equation $\left(y = \frac{3 - 3x}{2}\right)$:

It is easier to make $2x$ the subject to avoid fractions early on, or you can isolate $x$. Let's isolate $x$:

Alternatively, you could isolate $y = \frac{4 - 2x}{3}$, but substituting $x$ into $x^2$ is often less prone to sign errors than substituting into $-3y^2$. Let's stick with $x = \frac{4 - 3y}{2}$.

Replace $x$ in the first equation:

Square the fraction:

Multiply the entire equation by $4$ to remove the denominator:

Expand the bracket:

Combine like terms to form a quadratic equation equal to zero:

Divide the entire equation by $-3$ to simplify:

Factorise the quadratic equation:

Therefore, the solutions for $y$ are:

Substitute $y$ back into the rearranged linear equation $\left(x = \frac{4 - 3y}{2}\right)$.

When $y = -6$:

When $y = -2$: