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\( x+2 y=4 \) and \( 2 x-y-3=0 \) \( 2 y-5 x=7 \) and \( 4 x+3 y+24=0 \)

Ask by Malone Fernandez. in South Africa
Mar 10,2025

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Answer

The solutions are: 1. \( (x, y) = (2, 1) \) 2. \( (x, y) = (-3, -4) \)

Solution

**Solving the first system** We have: \[ x+2y=4 \quad \text{(1)} \] \[ 2x-y-3=0 \quad \text{(2)} \] 1. Rewrite equation (2) as: \[ 2x-y=3 \quad \text{(2')} \] 2. Solve (2') for \(y\): \[ y=2x-3 \] 3. Substitute \(y=2x-3\) into (1): \[ x+2(2x-3)=4 \] \[ x+4x-6=4 \] \[ 5x-6=4 \] 4. Solve for \(x\): \[ 5x=10 \quad \Rightarrow \quad x=2 \] 5. Substitute \(x=2\) back into \(y=2x-3\): \[ y=2(2)-3=4-3=1 \] Thus, the solution for the first system is: \[ \boxed{(x,y)=(2,1)} \] **Solving the second system** We have: \[ 2y-5x=7 \quad \text{(3)} \] \[ 4x+3y+24=0 \quad \text{(4)} \] 1. Rewrite equation (4) as: \[ 4x+3y=-24 \] 2. Solve (3) for \(y\): \[ 2y=5x+7 \quad \Rightarrow \quad y=\frac{5x+7}{2} \] 3. Substitute \(y=\frac{5x+7}{2}\) into equation (4): \[ 4x+3\left(\frac{5x+7}{2}\right)=-24 \] 4. Multiply the entire equation by 2 to eliminate the fraction: \[ 8x+3(5x+7)=-48 \] \[ 8x+15x+21=-48 \] \[ 23x+21=-48 \] 5. Solve for \(x\): \[ 23x=-48-21 = -69 \quad \Rightarrow \quad x=-3 \] 6. Substitute \(x=-3\) back into \(y=\frac{5x+7}{2}\): \[ y=\frac{5(-3)+7}{2}=\frac{-15+7}{2}=\frac{-8}{2}=-4 \] Thus, the solution for the second system is: \[ \boxed{(x,y)=(-3,-4)} \]

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Bonus Knowledge

To solve the first system of equations \( x + 2y = 4 \) and \( 2x - y - 3 = 0 \), we can express one variable in terms of the other and substitute. From the first equation, isolate \( y \): \( y = \frac{4 - x}{2} \). Substituting \( y \) into the second equation yields a solvable equation for \( x \). For the second system \( 2y - 5x = 7 \) and \( 4x + 3y + 24 = 0 \), similarly, isolate one variable. From the first equation, solve for \( y \): \( y = \frac{5x + 7}{2} \) and substitute back into the second equation to find \( x \). These steps lead you to the values of \( x \) and \( y \) for both systems of equations!

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