Answer :

To solve the simultaneous equations

[tex]\[ \begin{cases} 2x - y = 3 \\ x^2 - xy = -4 \end{cases} \][/tex]

we'll follow these steps:

1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from the first equation:

From the first equation:
[tex]\[ 2x - y = 3 \][/tex]

We can solve for [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 3 \][/tex]

2. Substitute [tex]\( y \)[/tex] in the second equation:

Now substitute [tex]\( y = 2x - 3 \)[/tex] into the second equation [tex]\( x^2 - xy = -4 \)[/tex]:
[tex]\[ x^2 - x(2x - 3) = -4 \][/tex]

3. Simplify and solve the resulting quadratic equation:

Simplify the equation:
[tex]\[ x^2 - 2x^2 + 3x = -4 \][/tex]
[tex]\[ -x^2 + 3x = -4 \][/tex]

Rearrange it into standard form:
[tex]\[ x^2 - 3x - 4 = 0 \][/tex]

4. Solve the quadratic equation:

We can solve the quadratic equation [tex]\( x^2 - 3x - 4 = 0 \)[/tex] by factoring:
[tex]\[ x^2 - 3x - 4 = (x + 1)(x - 4) = 0 \][/tex]

This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x + 1 = 0 \quad \text{or} \quad x - 4 = 0 \][/tex]
[tex]\[ x = -1 \quad \text{or} \quad x = 4 \][/tex]

5. Find the corresponding [tex]\( y \)[/tex] values:

For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 2(-1) - 3 \][/tex]
[tex]\[ y = -2 - 3 \][/tex]
[tex]\[ y = -5 \][/tex]

So, one solution is [tex]\( (x, y) = (-1, -5) \)[/tex].

For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 2(4) - 3 \][/tex]
[tex]\[ y = 8 - 3 \][/tex]
[tex]\[ y = 5 \][/tex]

So, the other solution is [tex]\( (x, y) = (4, 5) \)[/tex].

Therefore, the solutions to the system of equations are:
[tex]\[ (x, y) = (-1, -5) \quad \text{and} \quad (4, 5) \][/tex]

Other Questions