How do I solve these two equations simultaneously: 7x+y=1 and 2x^2 - y = 3

These types of questions can sometimes throw people off due to the quadratic nature of the second equation but don't fret; we'll use a method called substitution to find the answer!

Firstly, we rearrange the quadratic equation to leave the non-quadratic unknown on its own. Labelling the equations (1) and (2) will help with organisation later on

7x + y = 1  [1]     y = 2x^2 - 3  [2]

Next, we subsitute the quadratic expression into the other equation (i.e. replace y in [1] with the expression for y in [2]). Label this new equation as [3]

7x + (2x^2 - 3) = 1 [3]

Rearrange until in the form of a quadratic equation:

2x^2 + 7x - 3 - 1 = 0

2x^2 + 7x - 4  = 0

We now have a quadratic equation in x that needs to be solved (NOTE: we can use factorisation here!)

2x^2 + 7x - 4 = 0

(2x - 1)(x + 4) = 0

2x - 1 = 0        OR x + 4 =0

x = 0.5                   x  = -4 

So we have our x's! To find our y value we substitute our x values back into one of the original equations - we'll use [1]

when x = 0.5

[1] 7(0.5) + y = 1 

3.5 + y =1

y = 1 - 3.5 = -2.5

When x = -4

[1] 7(-4) + y = 1

-28  + y =1 

y = 1 + 28 = 29 We  now have our y-values for the respective x's!

We can check that this all adds up by substituting both pairs of answers back into the other original equation (in our case [2]).

So, for x = 0.5, y =-2.5

[2]  -2.5 = 2(0.5)^2 -3 

-2.5 = -2.5 (Check)

and for x = -4, y = 29

29 = 2(-4)^2 - 3

29 = 29 (Check)

Finally, be sure to write the answer pairs clearly at the bottom to grab those final marks for the correct answer!