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]
[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]