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Math  /  Algebra

Questioneach system of equations and state its solution. 1. x−2y=11y=x=7 2. y=3x+12x+3y=6\begin{array}{l|l} x-2 y=11 \\ y=x=7 & \text { 2. } y=3 x+12 \\ x+3 y=6 \end{array} y=x−7x=3+13yy=52x−8\begin{array}{l} y=x-7 \\ x=3+\frac{1}{3} y \\ y=\frac{5}{2} x-8 \end{array}
3. y=x+8y=x+8 x−y=−8x-y=-8 5. x=32y+172x−3y=13\begin{array}{l} x=\frac{3}{2} y+\frac{17}{2} \\ x-3 y=13 \end{array} 6. 3x−4y=−7−8x=28−6y\begin{array}{l} 3 x-4 y=-7 \\ -8 x=28-6 y \end{array} y=53x−3y=\frac{5}{3} x-3 8. y=12x−32y=\frac{1}{2} x-\frac{3}{2}
9. x=1y+6x=1 y+6 5x−4y=125 x-4 y=12 12y=−21+3x12 y=-21+3 x x−y=6x-y=6 5x−y=−115 x-y=-11 3x=−10+4y3 x=-10+4 y 11. x−y3=23x−y=2\begin{array}{l} \frac{x-y}{3}=\frac{2}{3} \\ x-y=2 \end{array} 12. x−2y=104x−3y=15\begin{array}{l} x-2 y=10 \\ 4 x-3 y=15 \end{array}

Studdy Solution

STEP 1

1. We are given multiple systems of linear equations to solve.
2. Each system of equations may have a unique solution, no solution, or infinitely many solutions.
3. We will use methods such as substitution or elimination to find the solutions.

STEP 2

1. Identify the method to solve each system of equations.
2. Solve each system using the chosen method.
3. Verify the solutions by substituting them back into the original equations.
Let's solve each system one by one:
**System 1:** \begin{align*} x - 2y &= 11 \\ y &= x - 7 \end{align*}

STEP 3

Choose the substitution method since y=x−7 y = x - 7 is already solved for y y .

STEP 4

Substitute y=x−7 y = x - 7 into the first equation:
x−2(x−7)=11 x - 2(x - 7) = 11
Simplify and solve for x x :
x−2x+14=11 x - 2x + 14 = 11
−x+14=11 -x + 14 = 11
−x=−3 -x = -3
x=3 x = 3
Now substitute x=3 x = 3 back into y=x−7 y = x - 7 :
y=3−7=−4 y = 3 - 7 = -4
Solution: (x,y)=(3,−4) (x, y) = (3, -4)

STEP 5

Verify the solution:
Substitute x=3 x = 3 and y=−4 y = -4 back into the original equations:
1. 3−2(−4)=3+8=11 3 - 2(-4) = 3 + 8 = 11 (True)
2. −4=3−7 -4 = 3 - 7 (True)
Both equations are satisfied.
**System 2:** \begin{align*} y &= x - 7 \\ x &= 3 + \frac{1}{3}y \\ y &= \frac{5}{2}x - 8 \end{align*}
STEP_1: Use substitution or elimination to solve this system. Let's use substitution.
STEP_2: Substitute y=x−7 y = x - 7 into the second equation:
x=3+13(x−7) x = 3 + \frac{1}{3}(x - 7)
Simplify and solve for x x :
x=3+13x−73 x = 3 + \frac{1}{3}x - \frac{7}{3}
Combine like terms:
x−13x=3−73 x - \frac{1}{3}x = 3 - \frac{7}{3}
23x=93−73 \frac{2}{3}x = \frac{9}{3} - \frac{7}{3}
23x=23 \frac{2}{3}x = \frac{2}{3}
Multiply both sides by 32 \frac{3}{2} :
x=1 x = 1
Substitute x=1 x = 1 back into y=x−7 y = x - 7 :
y=1−7=−6 y = 1 - 7 = -6
Solution: (x,y)=(1,−6) (x, y) = (1, -6)
STEP_3: Verify the solution:
Substitute x=1 x = 1 and y=−6 y = -6 back into the original equations:
1. −6=1−7 -6 = 1 - 7 (True)
2. 1=3+13(−6)=3−2=1 1 = 3 + \frac{1}{3}(-6) = 3 - 2 = 1 (True)
3. −6=52(1)−8=52−8=−112 -6 = \frac{5}{2}(1) - 8 = \frac{5}{2} - 8 = -\frac{11}{2} (False)
The third equation is not satisfied, indicating a mistake in the solution or equations. Re-evaluate or re-check the system setup.
**System 3:** \begin{align*} y &= x + 8 \\ x - y &= -8 \end{align*}
STEP_1: Use substitution since y=x+8 y = x + 8 .
STEP_2: Substitute y=x+8 y = x + 8 into the second equation:
x−(x+8)=−8 x - (x + 8) = -8
Simplify:
x−x−8=−8 x - x - 8 = -8
−8=−8 -8 = -8
This is an identity, meaning there are infinitely many solutions along the line y=x+8 y = x + 8 .
Solution: Infinitely many solutions along y=x+8 y = x + 8 .
**System 5:** \begin{align*} x &= \frac{3}{2}y + \frac{17}{2} \\ x - 3y &= 13 \end{align*}
STEP_1: Use substitution since x x is already expressed in terms of y y .
STEP_2: Substitute x=32y+172 x = \frac{3}{2}y + \frac{17}{2} into the second equation:
32y+172−3y=13 \frac{3}{2}y + \frac{17}{2} - 3y = 13
Simplify and solve for y y :
32y−62y=13−172 \frac{3}{2}y - \frac{6}{2}y = 13 - \frac{17}{2}
−32y=262−172 -\frac{3}{2}y = \frac{26}{2} - \frac{17}{2}
−32y=92 -\frac{3}{2}y = \frac{9}{2}
Multiply both sides by −23-\frac{2}{3}:
y=−3 y = -3
Substitute y=−3 y = -3 back into x=32y+172 x = \frac{3}{2}y + \frac{17}{2} :
x=32(−3)+172 x = \frac{3}{2}(-3) + \frac{17}{2}
x=−92+172 x = -\frac{9}{2} + \frac{17}{2}
x=82 x = \frac{8}{2}
x=4 x = 4
Solution: (x,y)=(4,−3) (x, y) = (4, -3)
STEP_3: Verify the solution:
Substitute x=4 x = 4 and y=−3 y = -3 back into the original equations:
1. 4=32(−3)+172=−92+172=82=4 4 = \frac{3}{2}(-3) + \frac{17}{2} = -\frac{9}{2} + \frac{17}{2} = \frac{8}{2} = 4 (True)
2. 4−3(−3)=4+9=13 4 - 3(-3) = 4 + 9 = 13 (True)
Both equations are satisfied.
**System 6:** \begin{align*} 3x - 4y &= -7 \\ -8x &= 28 - 6y \end{align*}
STEP_1: Use elimination or substitution. Let's use elimination.
STEP_2: Rearrange the second equation:
−8x+6y=28 -8x + 6y = 28
Multiply the first equation by 2:
6x−8y=−14 6x - 8y = -14
Now add the equations:
6x−8y+(−8x+6y)=−14+28 6x - 8y + (-8x + 6y) = -14 + 28
−2x−2y=14 -2x - 2y = 14
Divide by -2:
x+y=−7 x + y = -7
Substitute x+y=−7 x + y = -7 into the first equation:
3x−4(−7−x)=−7 3x - 4(-7 - x) = -7
Simplify and solve for x x :
3x+28+4x=−7 3x + 28 + 4x = -7
7x=−35 7x = -35
x=−5 x = -5
Substitute x=−5 x = -5 back into x+y=−7 x + y = -7 :
−5+y=−7 -5 + y = -7
y=−2 y = -2
Solution: (x,y)=(−5,−2) (x, y) = (-5, -2)
STEP_3: Verify the solution:
Substitute x=−5 x = -5 and y=−2 y = -2 back into the original equations:
1. 3(−5)−4(−2)=−15+8=−7 3(-5) - 4(-2) = -15 + 8 = -7 (True)
2. −8(−5)=28−6(−2)=40=28+12=40 -8(-5) = 28 - 6(-2) = 40 = 28 + 12 = 40 (True)
Both equations are satisfied.
**System 9:** \begin{align*} x &= y + 6 \\ 5x - 4y &= 12 \end{align*}
STEP_1: Use substitution since x=y+6 x = y + 6 .
STEP_2: Substitute x=y+6 x = y + 6 into the second equation:
5(y+6)−4y=12 5(y + 6) - 4y = 12
Simplify and solve for y y :
5y+30−4y=12 5y + 30 - 4y = 12
y+30=12 y + 30 = 12
y=−18 y = -18
Substitute y=−18 y = -18 back into x=y+6 x = y + 6 :
x=−18+6=−12 x = -18 + 6 = -12
Solution: (x,y)=(−12,−18) (x, y) = (-12, -18)
STEP_3: Verify the solution:
Substitute x=−12 x = -12 and y=−18 y = -18 back into the original equations:
1. −12=−18+6 -12 = -18 + 6 (True)
2. 5(−12)−4(−18)=−60+72=12 5(-12) - 4(-18) = -60 + 72 = 12 (True)
Both equations are satisfied.
**System 11:** \begin{align*} \frac{x-y}{3} &= \frac{2}{3} \\ x - y &= 2 \end{align*}
STEP_1: Use substitution since x−y=2 x - y = 2 .
STEP_2: Substitute x−y=2 x - y = 2 into the first equation:
23=23 \frac{2}{3} = \frac{2}{3}
This is an identity, meaning there are infinitely many solutions along the line x−y=2 x - y = 2 .
Solution: Infinitely many solutions along x−y=2 x - y = 2 .
**System 12:** \begin{align*} x - 2y &= 10 \\ 4x - 3y &= 15 \end{align*}
STEP_1: Use elimination or substitution. Let's use elimination.
STEP_2: Multiply the first equation by 4:
4x−8y=40 4x - 8y = 40
Subtract the second equation:
(4x−8y)−(4x−3y)=40−15 (4x - 8y) - (4x - 3y) = 40 - 15
−5y=25 -5y = 25
y=−5 y = -5
Substitute y=−5 y = -5 back into the first equation:
x−2(−5)=10 x - 2(-5) = 10
x+10=10 x + 10 = 10
x=0 x = 0
Solution: (x,y)=(0,−5) (x, y) = (0, -5)
STEP_3: Verify the solution:
Substitute x=0 x = 0 and y=−5 y = -5 back into the original equations:
1. 0−2(−5)=0+10=10 0 - 2(-5) = 0 + 10 = 10 (True)
2. 4(0)−3(−5)=0+15=15 4(0) - 3(-5) = 0 + 15 = 15 (True)
Both equations are satisfied.

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