-7=-3a+3b+17
how do you make the equation 1-3-5-7=10 true by adding perentheses?
x+y=4
$x^{5}$+ $y^{5}$=464
$\frac{4x}{10}$ + $\frac{x+5}{3}$ = $\frac{x}{5}$
∃𝑥 ∈ ℤ ∶ 2𝑥 + 1 = 4 ∨ 3𝑥 − 1 = −10
4a-3a when a=-2
4-(x+3)>3(3-x)
h-2k+5=0
2h+3k-1=0
-2(x+2)3(y-8)
5×3-2+1×2=