Linear and Quadratic Simultaneous Equations

Both equations give y, so the two expressions for y are equal.

1. Both equations give y, so set the two expressions equal: x^2-4=x+2.
2. Subtract x and subtract 2 from both sides: x^2-x-6=0.
3. The option x^2+x-6=0 comes from adding x instead of subtracting it, and x^2-x+6=0 has the wrong sign on the constant: -4-2=-6, not +6.

Substitute y=4 into y=x^2. Remember that x^2=4 has two solutions.

Part (a)

1. Substitution gives x^2=4.
2. Therefore x=-2 or x=2, and y=4 in both cases.
3. The smaller value of x gives (-2,4).

Part (b)

1. Substitution gives x^2=4.
2. Therefore x=-2 or x=2, and y=4 in both cases.
3. The larger value of x gives (2,4).

Both equations start y=\ldots, so set the two right-hand sides equal.

Part (a)

1. Equating the expressions gives x^2=x+6.
2. Rearrange and factorise: x^2-x-6=(x-3)(x+2)=0.
3. Thus x=-2 or x=3; substitute into y=x+6 to find the matching values.
4. For the smaller value x=-2, y=4, giving (-2,4).

Part (b)

1. Equating the expressions gives x^2=x+6.
2. Rearrange and factorise: x^2-x-6=(x-3)(x+2)=0.
3. Thus x=-2 or x=3; substitute into y=x+6 to find the matching values.
4. For the larger value x=3, y=9, giving (3,9).

Both equations start y=\ldots, so set the two right-hand sides equal and rearrange to \ldots=0.

Part (a)

1. Equating the expressions gives x^2-1=2x+2.
2. Rearrange and factorise: x^2-2x-3=(x-3)(x+1)=0.
3. Thus x=-1 or x=3. Find each matching y from y=2x+2.
4. The smaller x=-1 gives y=0, so the pair is (-1,0).

Part (b)

1. Equating the expressions gives x^2-1=2x+2.
2. Rearrange and factorise: x^2-2x-3=(x-3)(x+1)=0.
3. Thus x=-1 or x=3. Find each matching y from y=2x+2.
4. The larger x=3 gives y=8, so the pair is (3,8).

Set 5-x equal to x^2-1, then collect all the terms on one side.

Part (a)

1. Substitution gives 5-x=x^2-1.
2. This gives x^2+x-6=(x+3)(x-2)=0.
3. Thus x=-3 or x=2. Use y=5-x to find the matching y-values.
4. The smaller x=-3 gives y=8, so the pair is (-3,8).

Part (b)

1. Substitution gives 5-x=x^2-1.
2. This gives x^2+x-6=(x+3)(x-2)=0.
3. Thus x=-3 or x=2. Use y=5-x to find the matching y-values.
4. The larger x=2 gives y=3, so the pair is (2,3).

Rearrange x+y=7 to make y the subject, then substitute into xy=12.

Part (a)

1. The linear equation gives y=7-x.
2. Substitute into xy=12: x(7-x)=12.
3. Rearrange and factorise to get (x-3)(x-4)=0.
4. Thus x=3 or x=4, with y=7-x.
5. The smaller x=3 gives the pair (3,4).

Part (b)

1. The linear equation gives y=7-x.
2. Substitute into xy=12: x(7-x)=12.
3. Rearrange and factorise to get (x-3)(x-4)=0.
4. Thus x=3 or x=4, with y=7-x.
5. The larger x=4 gives the pair (4,3).

Make y the subject of x-y=3, and keep a bracket round it when you square it.

Part (a)

1. Rearrange the linear equation to make y the subject: y=x-3.
2. Substitute into x^2+y^2=45: x^2+(x-3)^2=45.
3. Expand the bracket: (x-3)^2=x^2-6x+9, so 2x^2-6x+9=45.
4. Rearrange to 2x^2-6x-36=0, then divide by 2: x^2-3x-18=0.
5. Factorise: (x-6)(x+3)=0, so x=-3 or x=6.
6. Find each y from y=x-3: x=-3 gives y=-6, and x=6 gives y=3.
7. The smaller value of x is -3, so the solution is (-3,-6).

Part (b)

1. Rearrange the linear equation to make y the subject: y=x-3.
2. Substitute into x^2+y^2=45: x^2+(x-3)^2=45.
3. Expand the bracket: (x-3)^2=x^2-6x+9, so 2x^2-6x+9=45.
4. Rearrange to 2x^2-6x-36=0, then divide by 2: x^2-3x-18=0.
5. Factorise: (x-6)(x+3)=0, so x=-3 or x=6.
6. Find each y from y=x-3: x=-3 gives y=-6, and x=6 gives y=3.
7. The larger value of x is 6, so the solution is (6,3).

Make y the subject of 2x+y=4, then substitute into xy=-6.

Part (a)

1. Rearrange 2x+y=4 to make y the subject: y=4-2x.
2. Substitute into xy=-6: x(4-2x)=-6, so 4x-2x^2=-6.
3. Rearrange to 2x^2-4x-6=0, then divide by 2: x^2-2x-3=0.
4. Factorise: (x-3)(x+1)=0, so x=-1 or x=3.
5. Find each y from y=4-2x: x=-1 gives y=6, and x=3 gives y=-2.
6. The smaller value of x is -1, so the solution is (-1,6).

Part (b)

1. Rearrange 2x+y=4 to make y the subject: y=4-2x.
2. Substitute into xy=-6: x(4-2x)=-6, so 4x-2x^2=-6.
3. Rearrange to 2x^2-4x-6=0, then divide by 2: x^2-2x-3=0.
4. Factorise: (x-3)(x+1)=0, so x=-1 or x=3.
5. Find each y from y=4-2x: x=-1 gives y=6, and x=3 gives y=-2.
6. The larger value of x is 3, so the solution is (3,-2).

Replace y with (x+1) in y^2=3x+7, keeping the bracket.

Part (a)

1. Substitution gives (x+1)^2=3x+7.
2. Expand and rearrange to get x^2-x-6=0.
3. Factorising gives (x-3)(x+2)=0, so x=-2 or x=3.
4. Use y=x+1 to find the y-value that goes with each x-value.
5. The smaller x=-2 gives the pair (-2,-1).

Part (b)

1. Substitution gives (x+1)^2=3x+7.
2. Expand and rearrange to get x^2-x-6=0.
3. Factorising gives (x-3)(x+2)=0, so x=-2 or x=3.
4. Use y=x+1 to find the y-value that goes with each x-value.
5. The larger x=3 gives the pair (3,4).

x is already the subject of the first equation, so replace x with 2y+1 in the second.

Part (a)

1. Substitute for x: y^2+2y+1=16.
2. Rearrange and factorise: y^2+2y-15=(y+5)(y-3)=0.
3. Thus y=-5 or y=3. Use x=2y+1 to find the matching x-values.
4. When y=-5, x=-9, giving the pair with smaller x, (-9,-5).

Part (b)

1. Substitute for x: y^2+2y+1=16.
2. Rearrange and factorise: y^2+2y-15=(y+5)(y-3)=0.
3. Thus y=-5 or y=3. Use x=2y+1 to find the matching x-values.
4. When y=3, x=7, giving the pair with larger x, (7,3).

Set x^2=4x-4 and rearrange. Check whether the quadratic is a perfect square.

1. Equate the expressions for y to get x^2-4x+4=0.
2. This is (x-2)^2=0, so there is just one value, x=2.
3. Substitute into y=x^2 to get y=4. The solution is (2,4).

Eliminate y, then decide whether the quadratic in x can ever equal 0.

1. Eliminate y by setting the expressions equal: x^2+2=x-2, so x^2-x+4=0.
2. Complete the square: x^2-x+4=\left(x-\frac12\right)^2+\frac{15}4.
3. A square is never negative, so \left(x-\frac12\right)^2+\frac{15}4 is always at least \frac{15}4. It can never equal 0.
4. So no value of x works, and the equations have no solutions: the line never meets the curve.

5. (The quadratic formula shows the same thing: b^2-4ac=(-1)^2-4\times1\times4=-15, and you cannot square-root a negative number.)

Make y the subject of x+y=10 and substitute, then use x>y to choose the pair.

1. Rearrange the linear equation: y=10-x.
2. Substitute into x^2+y^2=58: x^2+(10-x)^2=58.
3. Expand the bracket: (10-x)^2=100-20x+x^2, so 2x^2-20x+100=58.
4. Rearrange to 2x^2-20x+42=0, then divide by 2: x^2-10x+21=0.
5. Factorise: (x-3)(x-7)=0, so x=3 or x=7.
6. If x=3 then y=7; if x=7 then y=3. Only (7,3) has x>y.

Write one equation for the perimeter and one for the area, then substitute.

Part (a)

1. The perimeter gives 2x+2y=26, so x+y=13 and y=13-x.
2. The area gives xy=40. Substitute: x(13-x)=40, so 13x-x^2=40.
3. Rearrange: x^2-13x+40=0, which factorises to (x-5)(x-8)=0.
4. So x=5 or x=8, with y=13-x giving y=8 or y=5.
5. The length is the longer side, so x=8. The length is 8 cm.

Part (b)

1. The perimeter gives 2x+2y=26, so x+y=13 and y=13-x.
2. The area gives xy=40. Substitute: x(13-x)=40, so 13x-x^2=40.
3. Rearrange: x^2-13x+40=0, which factorises to (x-5)(x-8)=0.
4. So x=5 or x=8, with y=13-x giving y=8 or y=5.
5. The length is the longer side, so x=8 and y=13-8=5. The width is 5 cm.

Set t^2+2t equal to 5t+4, solve, then check which time is allowed.

1. Equating the distances gives t^2+2t=5t+4.
2. Rearrange and factorise: t^2-3t-4=(t-4)(t+1)=0.
3. The possible times are -1 and 4 seconds, but only 4 is in the model domain.
4. At t=4, d=5\times4+4=24 m. The answer is (4,24).

Make y the subject of 3x+y=4, then substitute it for y in x^2+2xy=-4.

Part (a)

1. Rearrange the linear equation: y=4-3x.
2. Substitute into x^2+2xy=-4: x^2+2x(4-3x)=-4.
3. Expand: x^2+8x-6x^2=-4, so -5x^2+8x+4=0.
4. Multiply by -1: 5x^2-8x-4=0. Factorise: (5x+2)(x-2)=0, so x=-\frac25 or x=2.
5. Find each y from y=4-3x: x=-\frac25 gives y=4+\frac65=\frac{26}5, and x=2 gives y=-2.
6. The smaller value of x is -\frac25, so the solution is \left(-\frac25,\frac{26}5\right).

Part (b)

1. Rearrange the linear equation: y=4-3x.
2. Substitute into x^2+2xy=-4: x^2+2x(4-3x)=-4.
3. Expand: x^2+8x-6x^2=-4, so -5x^2+8x+4=0.
4. Multiply by -1: 5x^2-8x-4=0. Factorise: (5x+2)(x-2)=0, so x=-\frac25 or x=2.
5. Find each y from y=4-3x: x=-\frac25 gives y=4+\frac65=\frac{26}5, and x=2 gives y=-2.
6. The larger value of x is 2, so the solution is (2,-2).

Make x the subject of x+2y=5, then substitute into 2xy+y^2=8.

Part (a)

1. Rearrange the linear equation to make x the subject: x=5-2y.
2. Substitute into 2xy+y^2=8: 2(5-2y)y+y^2=8.
3. Expand: 10y-4y^2+y^2=8, so 10y-3y^2=8.
4. Rearrange: 3y^2-10y+8=0. Factorise: (3y-4)(y-2)=0, so y=2 or y=\frac43.
5. Find each x from x=5-2y: y=2 gives x=1, and y=\frac43 gives x=5-\frac83=\frac73.
6. The smaller value of x is 1, so the solution is (1,2).

Part (b)

1. Rearrange the linear equation to make x the subject: x=5-2y.
2. Substitute into 2xy+y^2=8: 2(5-2y)y+y^2=8.
3. Expand: 10y-4y^2+y^2=8, so 10y-3y^2=8.
4. Rearrange: 3y^2-10y+8=0. Factorise: (3y-4)(y-2)=0, so y=2 or y=\frac43.
5. Find each x from x=5-2y: y=2 gives x=1, and y=\frac43 gives x=5-\frac83=\frac73.
6. The larger value of x is \frac73, so the solution is \left(\frac73,\frac43\right).

After substituting, the quadratic does not factorise: use the quadratic formula, and round only at the end.

Part (a)

1. Rearrange the linear equation: y=x-1.
2. Substitute into x^2+xy=14: x^2+x(x-1)=14, so 2x^2-x-14=0.
3. This does not factorise, so use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=2, b=-1, c=-14.
4. b^2-4ac=1+112=113, so x=\dfrac{1\pm\sqrt{113}}{4}, giving x=2.9075\ldots or x=-2.4075\ldots
5. Find each y from y=x-1, using the unrounded values: y=1.9075\ldots or y=-3.4075\ldots
6. Rounding to 2 decimal places, the solution with the smaller x is (-2.41,-3.41).

Part (b)

1. Rearrange the linear equation: y=x-1.
2. Substitute into x^2+xy=14: x^2+x(x-1)=14, so 2x^2-x-14=0.
3. This does not factorise, so use the quadratic formula x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=2, b=-1, c=-14.
4. b^2-4ac=1+112=113, so x=\dfrac{1\pm\sqrt{113}}{4}, giving x=2.9075\ldots or x=-2.4075\ldots
5. Find each y from y=x-1, using the unrounded values: y=1.9075\ldots or y=-3.4075\ldots
6. Rounding to 2 decimal places, the solution with the larger x is (2.91,1.91).

Substitute the line's expression for y into the equation of the circle.

Part (a)

1. At the points of intersection both equations are true, so substitute y=2x-4 into x^2+y^2=13: x^2+(2x-4)^2=13.
2. Expand the bracket: (2x-4)^2=4x^2-16x+16, so 5x^2-16x+16=13.
3. Rearrange: 5x^2-16x+3=0. Factorise: (5x-1)(x-3)=0, so x=\frac15 or x=3.
4. Find each y from y=2x-4: x=\frac15 gives y=\frac25-4=-\frac{18}5, and x=3 gives y=2.
5. The point with the smaller x is \left(\frac15,-\frac{18}5\right).

Part (b)

1. At the points of intersection both equations are true, so substitute y=2x-4 into x^2+y^2=13: x^2+(2x-4)^2=13.
2. Expand the bracket: (2x-4)^2=4x^2-16x+16, so 5x^2-16x+16=13.
3. Rearrange: 5x^2-16x+3=0. Factorise: (5x-1)(x-3)=0, so x=\frac15 or x=3.
4. Find each y from y=2x-4: x=\frac15 gives y=\frac25-4=-\frac{18}5, and x=3 gives y=2.
5. The point with the larger x is (3,2).

Multiply the first equation by 2 to clear the fraction, then make x the subject.

Part (a)

1. Multiply \dfrac x2+y=3 by 2 to clear the fraction: x+2y=6, so x=6-2y.
2. Substitute into x^2+xy+y^2=21: (6-2y)^2+(6-2y)y+y^2=21.
3. Expand each term: (6-2y)^2=36-24y+4y^2 and (6-2y)y=6y-2y^2.
4. Collect like terms: 3y^2-18y+36=21, so 3y^2-18y+15=0.
5. Divide by 3 and factorise: y^2-6y+5=(y-1)(y-5)=0, so y=1 or y=5.
6. Find each x from x=6-2y: y=1 gives x=4, and y=5 gives x=-4.
7. The smaller value of x is -4, so the solution is (-4,5).

Part (b)

1. Multiply \dfrac x2+y=3 by 2 to clear the fraction: x+2y=6, so x=6-2y.
2. Substitute into x^2+xy+y^2=21: (6-2y)^2+(6-2y)y+y^2=21.
3. Expand each term: (6-2y)^2=36-24y+4y^2 and (6-2y)y=6y-2y^2.
4. Collect like terms: 3y^2-18y+36=21, so 3y^2-18y+15=0.
5. Divide by 3 and factorise: y^2-6y+5=(y-1)(y-5)=0, so y=1 or y=5.
6. Find each x from x=6-2y: y=1 gives x=4, and y=5 gives x=-4.
7. The larger value of x is 4, so the solution is (4,1).