Here, we have two equations involving two variables, x and y.

The question asks us to 'solve' the set of simultaneous equations.

Equivalently, it is asking us to find the x values (and corresponding y values) which satisfy both of the equations.

We can approach this by substituion:

We will rearrange one equation so that one of the variables, for example y, is the subject of the equation (i.e it is on one side of the equation with only x terms on the other).

Looking at the two equations, the first one is already in this form.

So now we can substitute y in terms of x into the second equation, giving:

(x+3)² - x² + 3 = -6x

Now we have an equation in terms of one variable, x, which we should be able to solve.

Expand out the terms and group like terms so that we have the equation in a form that we know we can solve from:

Expanding:

x² + 3x + 3x + 9 - x² + 3 = -6x

Collecting like terms and simplyfying (x, x² and constants) on the Left:

6x + 12 = -6x

Bringing all terms to the LHS:

12x + 12 = 0

Factorising out the 12:

12(x + 1) = 0

Now, x + 1 = 0 as 12 doesn't equal zero

So, rearranging:

x = - 1

Substituing into the simpler equation for y (y = x + 3) gives:

y = - 1 + 3 = 2

We have now finished as we have the x and the corresponding y value that satisfy the simultaneous equations.

differentiate 2^x

Related Maths A Level answers

Solve the simultaneous equations y = x + 3, y^2 - x^2 + 3 = -6x

Prove or disprove the following statement: ‘No cube of an integer has 2 as its units digit.’

Given that Y=(x+3)(x+5); find dy/dx

solve the differential equation dy/dx=(3x*exp(4y))/(7+(2x^(2))^(2) when y = 0, x = 2