What is 5+2? Ive been totally stuck on this for the longest time. I really need help
1 answer:
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Start off by combining like terms on the LHS (the terms with x in them).
So we get
Replacing this result with what we had before on the LHS, we get
⇒Solve for x (divide both sides
)
⇒Don't forget about reciprocity rules when dividing. This is the same as multiplying both sides by
⇒
⇒
***This is a proper fraction
So we get
Replacing this result with what we had before on the LHS, we get
⇒Solve for x (divide both sides
⇒Don't forget about reciprocity rules when dividing. This is the same as multiplying both sides by
⇒
⇒
QUESTION 1
The given system of equations is
To solve by linear combination, we add equation (1) to equation (2) to get,
We divide through by 4 to obtain,
We put d=3 into equation (2) to get,
QUESTION 2
The given system is
4x + y = 5 ...eqn(1)
3x + y = 3 ...eqn(2)
To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.
This will give us,
This implies that,
Put x=3 into equation (1) to get,
The solution is
QUESTION 3
We want to solve the system;
a – 2b = –2 ....eqn(1)
2a + 2b = 14...eqn(2)
by linear combination.
We need to add equation (1) to equation (2) to eliminate b.
This implies that,
Simplify,
Divide both sides by 3 to get,
Put a=4 into equation (2) to obtain,
The ordered pair in the form (a, b) is
QUESTION 4
The given system of equations is
11x + 4y = 18 ...eqn(1)
3x + 4y = 2 ...eqn(2)
We subtract equation (2) from equation (1) to get,
Put x=2 into equation (2) to obtain,
This implies that,
The correct answer is (2,-1).
QUESTION 5
The given system is ;
2d + e = 8...eqn1
d – e = 4...eqn2
We add the two equations to eliminate e.
This implies that,
We divide both sides by 3 to get,
We put d=4 into equation (2) to get,
The solution is
The given system of equations is
To solve by linear combination, we add equation (1) to equation (2) to get,
We divide through by 4 to obtain,
We put d=3 into equation (2) to get,
QUESTION 2
The given system is
4x + y = 5 ...eqn(1)
3x + y = 3 ...eqn(2)
To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.
This will give us,
This implies that,
Put x=3 into equation (1) to get,
The solution is
QUESTION 3
We want to solve the system;
a – 2b = –2 ....eqn(1)
2a + 2b = 14...eqn(2)
by linear combination.
We need to add equation (1) to equation (2) to eliminate b.
This implies that,
Simplify,
Divide both sides by 3 to get,
Put a=4 into equation (2) to obtain,
The ordered pair in the form (a, b) is
QUESTION 4
The given system of equations is
11x + 4y = 18 ...eqn(1)
3x + 4y = 2 ...eqn(2)
We subtract equation (2) from equation (1) to get,
Put x=2 into equation (2) to obtain,
This implies that,
The correct answer is (2,-1).
QUESTION 5
The given system is ;
2d + e = 8...eqn1
d – e = 4...eqn2
We add the two equations to eliminate e.
This implies that,
We divide both sides by 3 to get,
We put d=4 into equation (2) to get,
The solution is
Caitlin has 27 $5 bills and 54 $10 bills.
Step-by-step explanation:
Given,
Worth of $5 and $10 bills = $675
Let,
Number of $5 bills = x
Number of $10 bills = y
According to given statement;
5x+10y=675 Eqn 1
She has twice as many $10 bills as $5 bills.
y=2x Eqn 2
Putting value of y from Eqn 2 in Eqn 1
Dividing both sides by 25
Putting x=27 in Eqn 2
Caitlin has 27 $5 bills and 54 $10 bills.
Keywords: linear equation, substitution method
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