Question

solve these systems of equations using any method. check each solution, if possible. homework help
1.
a. $2x + 3y = 9$
$-3x + 3y = -6$
b. $x = 8 - 2y$
$y - x = 4$
c. $y = -\frac{1}{2}x + 7$
$y = x - 8$
d. $9x + 10y = 14$
$7x + 5y = -3$

Explanation:

Part a

Step1: Subtract the two equations to eliminate \( y \)

We have the system:

$$
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$$

Subtract the second equation from the first: \((2x + 3y) - (-3x + 3y)=9 - (-6)\)
Simplify: \(2x + 3y + 3x - 3y = 9 + 6\)
Which gives: \(5x = 15\)

Step2: Solve for \( x \)

Divide both sides by 5: \(x=\frac{15}{5}=3\)

Step3: Substitute \( x = 3 \) into the first equation to find \( y \)

Substitute into \(2x + 3y = 9\): \(2(3)+3y = 9\)
Simplify: \(6 + 3y = 9\)
Subtract 6: \(3y = 9 - 6 = 3\)
Divide by 3: \(y = 1\)

Step4: Check the solution

Substitute \(x = 3\), \(y = 1\) into the second equation: \(-3(3)+3(1)=-9 + 3=-6\), which matches the second equation.

Step1: Substitute \( x = 8 - 2y \) into \( y - x = 4 \)

Substitute: \(y-(8 - 2y)=4\)
Simplify: \(y - 8 + 2y = 4\)
Combine like terms: \(3y - 8 = 4\)

Step2: Solve for \( y \)

Add 8: \(3y = 4 + 8 = 12\)
Divide by 3: \(y = 4\)

Step3: Substitute \( y = 4 \) into \( x = 8 - 2y \) to find \( x \)

Substitute: \(x = 8 - 2(4)=8 - 8 = 0\)

Step4: Check the solution

Substitute \(x = 0\), \(y = 4\) into \(y - x = 4\): \(4 - 0 = 4\), which is correct.

Step1: Set the two expressions for \( y \) equal

We have \(y = -\frac{1}{2}x + 7\) and \(y = x - 8\), so: \(-\frac{1}{2}x + 7 = x - 8\)

Step2: Solve for \( x \)

Add \(\frac{1}{2}x\) to both sides: \(7=\frac{3}{2}x - 8\)
Add 8: \(15=\frac{3}{2}x\)
Multiply both sides by \(\frac{2}{3}\): \(x = 15\times\frac{2}{3}=10\)

Step3: Substitute \( x = 10 \) into \( y = x - 8 \) to find \( y \)

Substitute: \(y = 10 - 8 = 2\)

Step4: Check the solution

Substitute \(x = 10\), \(y = 2\) into \(y = -\frac{1}{2}x + 7\): \(2=-\frac{1}{2}(10)+7=-5 + 7 = 2\), which is correct.

Answer:

\(x = 3\), \(y = 1\)

Part b