Solve the simultaneous equations. Show your working. \[ \left\{\begin{array}{l}4 x-3 y=28 \\ 6 x+y=9\end{array} \begin{array}{r}4 x-3 y=28 \\ 6 x+y=9\end{array}\right. \]
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Extra Insights
To solve the simultaneous equations, we'll use the substitution or elimination method. Let's start with the equations: 1. \( 4x - 3y = 28 \) (Equation 1) 2. \( 6x + y = 9 \) (Equation 2) We can solve Equation 2 for \( y \): \[ y = 9 - 6x \] Now, we substitute this expression for \( y \) into Equation 1: \[ 4x - 3(9 - 6x) = 28 \] Expanding this we get: \[ 4x - 27 + 18x = 28 \] Combining like terms: \[ 22x - 27 = 28 \] Adding 27 to both sides gives: \[ 22x = 55 \] Dividing both sides by 22, we find: \[ x = \frac{55}{22} = 2.5 \] Now we substitute \( x \) back into the equation for \( y \): \[ y = 9 - 6(2.5) \] \[ y = 9 - 15 \] \[ y = -6 \] Thus, the solution to the simultaneous equations is: \[ (x, y) = (2.5, -6) \] To verify, we can plug \( x \) and \( y \) back into both original equations: 1. Checking Equation 1: \[ 4(2.5) - 3(-6) = 10 + 18 = 28 \] ✓ 2. Checking Equation 2: \[ 6(2.5) + (-6) = 15 - 6 = 9 \] ✓ Both equations are satisfied, confirming that the solution is correct!
