Simultaneous Equations: past paper questions & worked solutions

Use both equations.

Double the second equation: \(4x-4y=-26\). Subtract this from the first: \(7y=35\), so \(y=5\).

Equal x coefficients allow elimination.

Substitute back to find the other unknown.

Use the two given equations.

Multiply the second equation by 4 to match the coefficients of x.

Subtract the first equation to eliminate x.

Substitute into the second original equation.

Add the equations: \[ 8x=20 \]

Substitute into \(5x+y=11\): \[ 5(2.5)+y=11 \]

Label the equations: $$\\begin{aligned} 5x + y &= 11 \\quad \\text{--- (1)} \\\\ 3x - y &= 9 \\quad \\text{--- (2)} \\end{aligned}$$

**Step 1: Add (1) and (2) to eliminate *y*.** $$\\begin{aligned} (5x + y) + (3x - y) &= 11 + 9 \\\\ 8x &= 20 \\\\ x &= \\frac{20}{8} = 2.5 \\end{aligned}$$

**Step 2: Substitute *x* = 2.5 into equation (1) to find *y*.** $$\\begin{aligned} 5(2.5) + y &= 11 \\\\ 12.5 + y &= 11 \\\\ y &= 11 - 12.5 = -1.5 \\end{aligned}$$

Subtract the second equation from the first: \[ 6y=-6 \]

Substitute into \(3x+2y=10\): \[ 3x+2(-1)=10 \]

From the second equation, \[ y=-3.5-4x \]

Substitute into the first equation: \[ 3x+5(-3.5-4x)=8 \]

Then \[ y=-3.5-4(-1.5)=2.5 \]

Solve: 3*x* + 2*y* = 10 and 3*x* − 4*y* = 16 algebraically.

Label the equations: $$\\begin{aligned} 3x + 2y &= 10 \\tag{1}\\\\ 3x - 4y &= 16 \\tag{2} \\end{aligned}$$

Subtract <span class="upright">(2)</span> from <span class="upright">(1)</span> to eliminate *x*: $$(1)-(2): \\quad 6y = -6 \\implies \\boxed{y = -1}$$

Substitute *y* =  − 1 into <span class="upright">(1)</span>: $$3x + 2(-1) = 10 \\implies 3x = 12 \\implies \\boxed{x = 4}$$

Check in equation <span class="upright">(2)</span>: 3(4) − 4(−1) = 12 + 4 = 16

From

Substitute into \(3x+5y=8\):

Then

Double the second equation: \[ 6x+10y=16 \]

Subtract the first equation: \[ 6y=15 \]

Substitute into \(3x+5y=8\): \[ 3x+5\left(\frac52\right)=8 \]

From the first equation, \[ y=x+2 \]

Substitute into the second equation: \[ 2(x+2)+x=1 \]

Then \[ y=-1+2=1 \]

Multiply the second equation by 3: \[ 21x+9y=24 \]

Subtract the first equation: \[ 19x=9.5 \]

Substitute into \(7x+3y=8\): \[ 7(0.5)+3y=8 \]

From the second equation: \[ y=3x-7 \]

Substitute into the first equation: \[ 7x+3(3x-7)=3 \]

Then \[ y=3\left(\frac32\right)-7=\frac92-7=-\frac52 \]

Double the first equation: \[ 6x+10y=6.2 \]

Subtract the second equation: \[ 7y=2.45 \]

Substitute into \(3x+5y=3.1\): \[ 3x+5(0.35)=3.1 \]

Double the first equation: \[ 10a+4c=20 \]

Add this to the second equation: \[ 12a=27 \]

Substitute into \(5a+2c=10\): \[ 5\left(\frac94\right)+2c=10 \]

Add the equations: \[ 10x=-5 \]

Substitute into \(3x+5y=6\): \[ 3\left(-\frac12\right)+5y=6 \]

From

Substitute into \(7x+3y=3\):

Then

Multiply the first equation by 5: \[ 20x+15y=48 \]

Multiply the second equation by 3: \[ 18x+15y=50.4 \]

Subtract: \[ 2x=-2.4 \]

Substitute into \(4x+3y=9.6\): \[ 4(-1.2)+3y=9.6 \]

Multiply the first equation by 3: \[ 9x-15y=75 \]

Multiply the second equation by 5: \[ 20x+15y=70 \]

Add: \[ 29x=145 \]

Substitute into \(3x-5y=25\): \[ 15-5y=25 \]

Multiply the first equation by 5: \[ 10x+35y=85 \]

Multiply the second equation by 2: \[ 10x+6y=-2 \]

Subtract: \[ 29y=87 \]

Substitute into \(2x+7y=17\): \[ 2x+21=17 \]

Multiply the first equation by 5: \[ 35x-10y=170 \]

Multiply the second equation by 2: \[ 6x+10y=-6 \]

Add: \[ 41x=164 \]

Substitute into \(3x+5y=-3\): \[ 12+5y=-3 \]

Multiply the first equation by 5: \[ 35x+10y=27.5 \]

Multiply the second equation by 2: \[ 6x-10y=34 \]

Add: \[ 41x=61.5 \]

Substitute into \(7x+2y=5.5\): \[ 10.5+2y=5.5 \]

\(x=3\Rightarrow y=4\);\quad \(x=\tfrac34\Rightarrow y=\tfrac94-5=-\tfrac{11}{4}\)

Substitute the linear expression for y into the nonlinear equation.

Collect all terms on one side and divide by 2.

Factorize the quadratic and set each factor to zero.

For \(x=\frac34\), \(y=3(\frac34)-5=-\frac{11}{4}\). For \(x=3\), \(y=9-5=4\).

Find the corresponding y-value for each x, preserving the pairs.

Substitute \(y=3x-2\) into \(x^2+y^2-2y=12\): \[x^2+(3x-2)^2-2(3x-2)=12.\]

Both equations must hold at an intersection.

Expand carefully, collect like terms, and divide by 2.

Use factorisation and the zero-product property.

When \(x=-\tfrac15\), \(y=3(-\tfrac15)-2=-\tfrac{13}{5}\). When \(x=2\), \(y=3(2)-2=4\).

Substitute each root into the line, preserving the coordinate pairings.

**Step 1: Substitute *y* = (*x*+2) into the quadratic equation.** *x*² + (*x*+2)² = 3*x*(*x*+2) − 11

*x*² + (*x*²+4*x*+4) = 3*x*² + 6*x* − 11 2*x*² + 4*x* + 4 = 3*x*² + 6*x* − 11

0 = 3*x*² − 2*x*² + 6*x* − 4*x* − 11 − 4 *x*² + 2*x* − 15 = 0

(*x*+5)(*x*−3) = 0 So, *x* =  − 5 or *x* = 3.

**Step 5: Find corresponding *y* values using *y* = *x* + 2.**

If *x* =  − 5 ⟹ *y* =  − 5 + 2 =  − 3

If *x* = 3 ⟹ *y* = 3 + 2 = 5

Substitute:

So

Find \(y\):

So

Substitute:

So

Find \(y\):

Solve *y*² + 5*y* + *x*² = 12 and *x* = *y* + 5.

Substitute *x* = *y* + 5 into the first equation: *y*² + 5*y* + (*y*+5)² = 12

Expand (*y*+5)² = *y*² + 10*y* + 25: *y*² + 5*y* + *y*² + 10*y* + 25 = 12 ⟹ 2*y*² + 15*y* + 13 = 0

Factorise: $$(2y+13)(y+1) = 0 \\implies y = -\\tfrac{13}{2}\\;\\text{ or }\\;y = -1$$

Find corresponding *x* values using *x* = *y* + 5:

Substitute:

So

Find \(y\):

Substitute:

So

Find \(y\):

Substitute:

So

Find \(x\):

So

Substitute into

So

Find \(x\):

Substitute \(y=2x-3\): \[ 3x^2+(2x-3)^2-x(2x-3)=5 \]

Factorise: \[ (5x-4)(x-1)=0 \]

So \[ x=\frac45 \quad \text{or} \quad x=1 \]

If \(x=\frac45\): \[ y=2\left(\frac45\right)-3=-\frac75 \]

If \(x=1\): \[ y=2(1)-3=-1 \]

Let \(a\) be the cost of one apple and \(p\) be the cost of one pear.

Multiply the first equation by 2:

Multiply the second equation by 3:

Subtract:

Then

Michael buys 10 apples and 10 pears:

Substitute \(y=3-2x\): \[ x^2+(3-2x)^2=18 \]

Factorise: \[ (5x+3)(x-3)=0 \]

So \[ x=-\frac35 \quad \text{or} \quad x=3 \]

If \(x=-\frac35\), then \[ y=3-2\left(-\frac35\right)=\frac{21}{5} \]

If \(x=3\), then \[ y=3-6=-3 \]

From the first equation: \[ x=6y+5 \]

Substitute into the second equation: \[ y(6y+5)-2y^2=6 \]

Factorise: \[ (4y-3)(y+2)=0 \]

So \[ y=\frac34 \quad \text{or} \quad y=-2 \]

If \(y=\frac34\): \[ x=6\left(\frac34\right)+5=\frac{19}{2} \]

If \(y=-2\): \[ x=6(-2)+5=-7 \]

From the second equation: \[ x=1-2y \]

Substitute into the first equation: \[ (1-2y)^2-9y-(1-2y)=2y^2-12 \]

Factorise: \[ (2y-3)(y-4)=0 \]

So \[ y=\frac32 \quad \text{or} \quad y=4 \]

If \(y=\frac32\): \[ x=1-2\left(\frac32\right)=-2 \]

If \(y=4\): \[ x=1-8=-7 \]

From

Substitute into \(x^2-y^2+2x=10\):

So

If \(y=-\frac13\),

If \(y=-5\),