There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10
[2] b-c=15
[3] a-2b+c=-5
Equations Simplified or Rearranged :<span><span> [1] 2a + c = -10
</span><span> [2] - c + b = 15
</span><span> [3] a + c - 2b = -5
</span></span>Solve by Substitution :
// Solve equation [3] for the variable c
<span> [3] c = -a + 2b - 5
</span>
// Plug this in for variable c in equation [1]
<span><span> [1] 2a + (-a +2?-5) = -10
</span><span> [1] a = -5
</span></span>
// Plug this in for variable c in equation [2]
<span><span> [2] - (-? +2b-5) + b = 15
</span><span> [2] - b = 10
</span></span>
// Solve equation [2] for the variable ?
<span> [2] ? = b + 10
</span>
// Plug this in for variable ? in equation [1]
<span><span> [1] (? +10) = -5
</span><span> [1] 0 = -15 => NO solution
</span></span><span>No solution</span>
Answer:
x= 5
Step-by-step explanation:
first we open brackets
12x-8=52
then transpose
12x= 52+8
12x= 60
x= 60/12
x= 5
hope it helps , pls mark me as brainliest
Answer with Step-by-step explanation:
We are given that
For each real number 
To prove that f is one -to-one.
Proof:Let 
and 
be any nonzero real numbers such that
By using the definition of f to rewrite the left hand side of this equation
Then, by using the definition of f to rewrite the right hand side of this equation of
Equating the expression then we get
Therefore, f is one-to-one.
The second attachment I solved in your another question.You may refer to that.
#1
Apply Pythagorean theorem
x²=10²-6²
