Answer
Distance between x-intercepts is greater by 6 units
Work Step by Step
Let $x=0$ (y-intercepts)
$x^2/100 + y^2/49 =1$
$0^2/100 + y^2/49 =1$
$0/100 +y^2/49 =1$
$y^2/49 =1$
$y^2*49/49=1*49$
$y^2=49$
$\sqrt{y^2} = \sqrt {49}$
$y=±7$
$7- (-7)=14$
Let $y=0$ (x-intercepts)
$x^2/100 + y^2/49 =1$
$x^2/100 + 0^2/49 =1$
$x^2/100 +0 = 1$
$x^2/100 =1$
$x^2/100*100 =1*100$
$x^2=100$
$\sqrt {x^2} =\sqrt {100}$
$x=±10$
$10-(-10)$
$20$
