Algebra
Calculus
Trigonometry
Matrix
Question
Solve the system of equations
Solve using the substitution method
Solve using the elimination method
(a1,b1)=(105+35,10−5+35)(a2,b2)=(105−35,−105+35)
Evaluate
{a−b=5ab5ab=1
Solve the equation for a
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Evaluate
a−b=5ab
Evaluate
a−b=5ba
Move the variable to the left side
a−b−5ba=0
Collect like terms by calculating the sum or difference of their coefficients
(1−5b)a−b=0
Move the constant to the right side
(1−5b)a=0+b
Removing 0 doesn't change the value,so remove it from the expression
(1−5b)a=b
Divide both sides
1−5b(1−5b)a=1−5bb
Divide the numbers
a=1−5bb
{a=1−5bb5ab=1
Substitute the given value of a into the equation 5ab=1
5×1−5bb×b=1
Simplify
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Evaluate
5×1−5bb×b
Multiply the terms
1−5b5b×b
Multiply the terms
1−5b5b×b
Multiply the terms
1−5b5b2
1−5b5b2=1
Cross multiply
5b2=1−5b
Move the expression to the left side
5b2−(1−5b)=0
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
5b2−1+5b=0
Rewrite in standard form
5b2+5b−1=0
Substitute a=5,b=5 and c=−1 into the quadratic formula b=2a−b±b2−4ac
b=2×5−5±52−4×5(−1)
Simplify the expression
b=10−5±52−4×5(−1)
Simplify the expression
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Evaluate
52−4×5(−1)
Multiply
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Multiply the terms
4×5(−1)
Any expression multiplied by 1 remains the same
−4×5
Multiply the terms
−20
52−(−20)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
52+20
Evaluate the power
25+20
Add the numbers
45
b=10−5±45
Simplify the radical expression
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Evaluate
45
Write the expression as a product where the root of one of the factors can be evaluated
9×5
Write the number in exponential form with the base of 3
32×5
The root of a product is equal to the product of the roots of each factor
32×5
Reduce the index of the radical and exponent with 2
35
b=10−5±35
Separate the equation into 2 possible cases
b=10−5+35b=10−5−35
Use b−a=−ba=−ba to rewrite the fraction
b=10−5+35b=−105+35
Evaluate the logic
b=10−5+35∪b=−105+35
Rearrange the terms
{a=1−5bbb=10−5+35∪{a=1−5bbb=−105+35
Calculate
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Evaluate
{a=1−5bbb=10−5+35
Substitute the given value of b into the equation a=1−5bb
a=1−5×10−5+3510−5+35
Calculate
a=105+35
Calculate
{a=105+35b=10−5+35
{a=105+35b=10−5+35∪{a=1−5bbb=−105+35
Calculate
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Evaluate
{a=1−5bbb=−105+35
Substitute the given value of b into the equation a=1−5bb
a=1−5(−105+35)−105+35
Simplify the expression
a=−1−5(−105+35)105+35
Calculate
a=105−35
Calculate
{a=105−35b=−105+35
{a=105+35b=10−5+35∪{a=105−35b=−105+35
Check the solution
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Check the solution
⎩⎨⎧105+35−10−5+35=5(105+35×10−5+35)5(105+35×10−5+35)=1
Simplify
{1=11=1
Evaluate
true
{a=105+35b=10−5+35∪{a=105−35b=−105+35
Check the solution
More Steps
Check the solution
⎩⎨⎧105−35−(−105+35)=5(105−35×(−105+35))5(105−35×(−105+35))=1
Simplify
{1=11=1
Evaluate
true
{a=105+35b=10−5+35∪{a=105−35b=−105+35
Solution
(a1,b1)=(105+35,10−5+35)(a2,b2)=(105−35,−105+35)
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