Eliminate the value of
2x+3y+4z=29
3x-2y+5z=20
4x+5y+3z=35
Answers
Answer:
xx
4y−2x+ 5y+2x9y+0=4=23=27
xx
9y+09y y=27=27=3Divide each side by 9
xxyy33
\begin{aligned} 5\blueD y + 2x &= 23&\gray{\text{Equation 2}} \\\\ 5(\blueD 3) + 2x &= 23 &\gray{\text{Substitute 3 for y}}\\\\ 15 + 2x &= 23 &\gray{\text{}}\\\\ 2x &= 8 &\gray{\text{Subtract 15 from each side}}\\\\ \greenD x &\greenD = \greenD 4&\gray{\text{Divide each side by 4}}\end{aligned}
Multiplying one of the equations by a constant, then using elimination
That last example worked out great because the yyy variable was eliminated when we added the equations. Sometimes it isn't quite that easy.
Take this system of equation as an example:
6x + 5y = 28~~~~~~~~ \gray{\text{Equation 1}}6x+5y=28 Equation 16, x, plus, 5, y, equals, 28, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
3x - 4y = 1~~~~~~~~ \gray{\text{Equation 2}}3x−4y=1 Equation 23, x, minus, 4, y, equals, 1, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
If we add these equations, neither the xxx or yyy variable will be eliminated, so that won't work. Here are the steps for problems like this:
Step 1: Multiply one of the equations by a constant so that when we add it to the other equation, one of the variables is eliminated.
\begin{aligned} \maroonD{-2}(3x -4y) &= \maroonD{-2}(1) &\gray{\text{Multiply the second equation by} -2} \\\\ \blueD{-6x+8y} &\blueD= \blueD{-2}&\gray{\text{Simplify to get a new equation}}\end{aligned}
−2(3x−4y)
−6x+8y
=−2(1)
=−2
Multiply the second equation by−2
Simplify to get a new equation
Step 2: Add the new equation to the equation we didn't use in step 1 in order to eliminate one of the variables.
6x+5y+ −6x+8y13y=28=−2=26Equation 1The new equation
Step 3: Solve for yyy.
\begin{aligned} 13y &= 26\\\\ y&= 2&\gray{\text{Divide each side by 13}}\end{aligned}
13y
y
=26
=2
Divide each side by 13
Step 4: Substitute y = 2y=2y, equals, 2 into one of the original equations, and solve for xxx.
\begin{aligned} 3x - 4y &= 1 &\gray{\text{Equation 2}} \\\\ 3x -4(2) &= 1 &\gray{\text{Substitute 2 for y}} \\\\ 3x -8 &= 1 \\\\ 3x &= 9 &\gray{\text{Add 8 to each side}} \\\\ x &= 3 &\gray{\text{Divide each side by 3}} \end{aligned}
3x−4y
3x−4(2)
3x−8
3x
x
=1
=1
=1
=9
=3
Equation 2
Substitute 2 for y
Add 8 to each side
Divide each side by 3
So our solution is (3, 2)(3,2)left parenthesis, 3, comma, 2, right parenthesis.
Use elimination to solve the following system of equations.
8x + 14y = 128x+14y=128, x, plus, 14, y, equals, 12
-6x - 7y = -16−6x−7y=−16minus, 6, x, minus, 7, y, equals, minus, 16
x =x=x, equals
y =y=y, equals
Show solution
22
8x + 14y = 128, x, plus, 14, y, equals, 12
-12x - 14y = -32minus, 12, x, minus, 14, y, equals, minus, 32
-4x = -20minus, 4, x, equals, minus, 20
xx
x = 5x, equals, 5
yy
8x + 14y = 128, x, plus, 14, y, equals, 12
8(5) + 14y = 128, left parenthesis, 5, right parenthesis, plus, 14, y, equals, 12
40 + 14y = 1240, plus, 14, y, equals, 12
14y = -2814, y, equals, minus, 28
y = -2y, equals, minus, 2
x = 5x, equals, 5
y = -2y, equals, minus, 2
Multiplying both of the equations by a constant, then using elimination
Sometimes we'll need to multiply both equations by a constant when using elimination.
For example, consider this system of equations:
5x + 3y = 14 ~~~~~~~~ \gray{\text{Equation 1}}5x+3y=14 Equation 15, x, plus, 3, y, equals, 14, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
3x + 2y = 8 ~~~~~~~~ \gray{\text{Equation 2}}3x+2y=8 Equation 23, x, plus, 2, y, equals, 8, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
Here are the steps to solve a system of equations like this one:
Step 1: Multiply each equation by a constant so we can eliminate one variable.
5x + 3y = 14 \Rightarrow \maroonD{\text{Multiply by 2}} \Rightarrow 10x + 6y = 285x+3y=14⇒Multiply by 2⇒10x+6y=285, x, plus, 3, y, equals, 14, right arrow, start color #ca337c, start text, M, u, l, t, i, p, l, y, space, b, y, space, 2, end text, end color #ca337c, right arrow, 10, x, plus, 6, y, equals, 28
3x + 2y = 8 \Rightarrow \maroonD{\text{Multiply by 3}} \Rightarrow 9x + 6y = 243x+2y=8⇒Multiply by 3⇒9x+6y=24
Answer:
xx
4y−2x+ 5y+2x9y+0=4=23=27
xx
9y+09y y=27=27=3Divide each side by 9
xxyy33
\begin{aligned} 5\blueD y + 2x &= 23&\gray{\text{Equation 2}} \\\\ 5(\blueD 3) + 2x &= 23 &\gray{\text{Substitute 3 for y}}\\\\ 15 + 2x &= 23 &\gray{\text{}}\\\\ 2x &= 8 &\gray{\text{Subtract 15 from each side}}\\\\ \greenD x &\greenD = \greenD 4&\gray{\text{Divide each side by 4}}\end{aligned}
Multiplying one of the equations by a constant, then using elimination
That last example worked out great because the yyy variable was eliminated when we added the equations. Sometimes it isn't quite that easy.
Take this system of equation as an example:
6x + 5y = 28~~~~~~~~ \gray{\text{Equation 1}}6x+5y=28 Equation 16, x, plus, 5, y, equals, 28, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
3x - 4y = 1~~~~~~~~ \gray{\text{Equation 2}}3x−4y=1 Equation 23, x, minus, 4, y, equals, 1, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
If we add these equations, neither the xxx or yyy variable will be eliminated, so that won't work. Here are the steps for problems like this:
Step 1: Multiply one of the equations by a constant so that when we add it to the other equation, one of the variables is eliminated.
\begin{aligned} \maroonD{-2}(3x -4y) &= \maroonD{-2}(1) &\gray{\text{Multiply the second equation by} -2} \\\\ \blueD{-6x+8y} &\blueD= \blueD{-2}&\gray{\text{Simplify to get a new equation}}\end{aligned}
−2(3x−4y)
−6x+8y
=−2(1)
=−2
Multiply the second equation by−2
Simplify to get a new equation
Step 2: Add the new equation to the equation we didn't use in step 1 in order to eliminate one of the variables.
6x+5y+ −6x+8y13y=28=−2=26Equation 1The new equation
Step 3: Solve for yyy.
\begin{aligned} 13y &= 26\\\\ y&= 2&\gray{\text{Divide each side by 13}}\end{aligned}
13y
y
=26
=2
Divide each side by 13
Step 4: Substitute y = 2y=2y, equals, 2 into one of the original equations, nd solve for xxx.
\begin{aligned} 3x - 4y &= 1 &\gray{\text{Equation 2}} \\\\ 3x -4(2) &= 1 \gray{\text{Substitute 2 for y}} \\\\ 3x -8 &= 1 \\\\ 3x &= 9 &\gray{\text{Add 8 to ach side}} \\\\ x &= 3 &\gray{\text{Divide each side by 3}} \end{aligned}
3x−4y
3x−4(2)
3x−8
3x
x
=1
=1
=1
=9
=3
Equation 2
Substitute 2 for y
Add 8 to each side
Divide each side by 3
So our solution is (3, 2)(3,2)left parenthesis, 3, comma, 2, right parenthesis.
Use elimination to solve the following system of equations.
8x + 14y = 128x+14y=128, x, plus, 14, y, equals, 12
-6x - 7y = -16−6x−7y=−16minus, 6, x, minus, 7, y, equals, minus, 16
x =x=x, equals
y =y=y, equals
Show solution
22
8x + 14y = 128, x, plus, 14, y, equals, 12
-12x - 14y = -32minus, 12, x, minus, 14, y, equals, minus, 32
-4x = -20minus, 4, x, equals, minus, 20
xx
x = 5x, equals, 5
yy
8x + 14y = 128, x, plus, 14, y, equals, 12
8(5) + 14y = 128, left parenthesis, 5, right parenthesis, plus, 14, y, equals, 12
40 + 14y = 1240, plus, 14, y, equals, 12
14y = -2814, y, equals, minus, 28
y = -2y, equals, minus, 2
x = 5x, equals, 5
y = -2y, equals, minus, 2
Multiplying both of the equations by a constant, then using elimination
Sometimes we'll need to multiply both equations by a constant when using elimination.
For example, consider this system of equations:
5x + 3y = 14 ~~~~~~~~ \gray{\text{Equation 1}}5x+3y=14 Equation 15, x, plus, 3, y, equals, 14, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
3x + 2y = 8 ~~~~~~~~ \gray{\text{Equation 2}}3x+2y=8 Equation 23, x, plus, 2, y, equals, 8, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
Here are the steps to solve a system of equations like this one:
Step 1: Multiply each equation by a constant so we can eliminate one variable.
5x + 3y = 14 \Rightarrow \maroonD{\text{Multiply by 2}} \Rightarrow 10x + 6y = 285x+3y=14⇒Multiply by 2⇒10x+6y=285, x, plus, 3, y, equals, 14, right arrow, start color #ca337c, start text, M, u, l, t, i, p, l, y, space, b, y, space, 2, end text, end color #ca337c, right arrow, 10, x, plus, 6, y, equals, 28
3x + 2y = 8 \Rightarrow \maroonD{\text{Multiply by 3}} \Rightarrow 9x + 6y = 243x+2y=8⇒Multiply by 3⇒9x+6y=24