38+6(y+1)= 60 find y
Answers
Answer:
Key idea: Anytime we have two true equations, we can add or subtract them to create another true equation.
For example, here are two very basic true equations:
2 = 22=22, equals, 2
5 = 55=55, equals, 5
We can add these equations together to create another true equation:
2+ 57=2=5=7
Or we can subtract these equations to create another true equation:
2− 5−3=2=5=−3
Here's another example with more complicated equations:
2x+3+ 4x+16x+4=7=9=16
Great, now that we see that it's okay to add or subtract equations, we can move onto solving a system of equations using elimination.
Solving a system of equations using elimination
We'll solve this system of equations as an example:
x + 3y = 8~~~~~~~~ \gray{\text{Equation 1}}x+3y=8 Equation 1x, plus, 3, 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, 1, end text, end color gray
4x - 3y = 17~~~~~~~~\gray{\text{Equation 2}}4x−3y=17 Equation 24, x, minus, 3, y, equals, 17, 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
The hard thing about solving is that there are two variables xxx and yyy. If only we could get rid of one of the variables...
Here's an idea! Let's add the two equations together to cancel out the yyy variable:
x+3y+ 4x−3y5x+0=8=17=25
Brilliant! Now we have an equation with just the xxx variable that we know how to solve:
5x+05x x=25=25=5Divide each side by 5
Baller! Let's use the first equation to find yyy when xxx equals 555:
\begin{aligned} \blueD x + 3y &= 8&\gray{\text{Equation 1}} \\\\ \blueD 5 + 3y &= 8 &\gray{\text{Substitute 5 for x}}\\\\ 3y &= 3 &\gray{\text{Subtract 5 from each side}}\\\\ \greenD y &\greenD = \greenD 1&\gray{\text{Divide each side by 3}}\end{aligned}
x+3y
5+3y
3y
y
=8
=8
=3
=1
Equation 1
Substitute 5 for x
Subtract 5 from each side
Divide each side by 3
Sweet! So the solution to the system of equations is (\blueD5, \greenD{1})(5,1)left parenthesis, start color #11accd, 5, end color #11accd, comma, start color #1fab54, 1, end color #1fab54, right parenthesis. [Let's check this solution in the original equations!]
Use elimination to solve the following system of equations.
4y - 2x = 44y−2x=44, y, minus, 2, x, equals, 4
5y + 2x = 235y+2x=235, y, plus, 2, x, equals, 23
x =x=x, equals
y =y=y, equals
[Show solution]
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
Step-by-step explanation:
Answer:
y is equals to 8/3
Step-by-step explanation:
this is your final answer