Question

1. find the solution to the following systems of linear equations by using the elimination method. \\(\

$$\begin{cases}9x + 4y = 9 \\\\ 8x + 3y = 8\\end{cases}$$

\\)

Explanation:

Step1: Make coefficients of y equal

Multiply the first equation by 3 and the second equation by 4 to make the coefficients of \( y \) equal.
First equation: \( 3\times(9x + 4y)=3\times9 \) gives \( 27x+12y = 27 \)
Second equation: \( 4\times(8x + 3y)=4\times8 \) gives \( 32x+12y = 32 \)

Step2: Eliminate y by subtraction

Subtract the first new equation from the second new equation to eliminate \( y \).
\( (32x + 12y)-(27x + 12y)=32 - 27 \)
Simplify: \( 32x+12y - 27x - 12y=5 \)
Which gives \( 5x=5 \)

Step3: Solve for x

Divide both sides by 5: \( x = \frac{5}{5}=1 \)

Step4: Substitute x into an original equation

Substitute \( x = 1 \) into the first original equation \( 9x+4y = 9 \)
\( 9\times1+4y = 9 \)
\( 9 + 4y=9 \)

Step5: Solve for y

Subtract 9 from both sides: \( 4y=9 - 9=0 \)
Divide by 4: \( y=\frac{0}{4}=0 \)

Answer:

The solution is \( x = 1 \) and \( y = 0 \) (or the ordered pair \( (1,0) \))